Measurement

Being able to solve problems involving surface area and volume in both SI (metric) and imperial units is important in fields like engineering, construction and daily life. In this section, we will review common units used for surface area measurements. In the SI system, we use square metres (m²), while in the imperial system, square feet (ft²) or square inches (in²) are used. Understanding these units and how to convert them is essential for accurate calculations.

We will also focus on solving real-world problems, like determining the amount of paint needed to cover a wall or the material required to wrap a gift. Whether we are using square metres or square feet, knowing how to convert between units is important.

For volume, we will work with SI units like cubic metre (m³) or litres (l), and imperial units like cubic feet (ft³) or gallons. Mastering these skills helps us solve practical problems in both school and everyday life.

KEY IDEAS

• Imperial Units are measurements historically used in the British Empire and still in use in some countries like the United States. Even in Ghana, carpenters use feet and inches while traders use pounds and gallons.

Examples include feet (ft), pounds (lbs) and gallons (gal) for length, weight, and volume, respectively.

• SI Unit is the International System of Units (SI). It is the standard system of measurement used globally. It includes units like the metre (m) for length, kilogram (kg) for mass and litre (l) for volume.

• Surface Area is the total area of the outer surface of a 3-D (three dimensional) object. It is measured in square units. It represents the amount of material needed to cover the surface of an object, like the surface of a cube or a sphere.

• Unit Conversion is the process of changing a measurement from one unit to another, such as converting from centimetres to inches or from litres to gallons. This often requires multiplying or dividing by a conversion factor.

• Volume/Capacity is the amount of space a 3-D object occupies, measured in cubic units. Volume refers to the interior space of objects like cubes or spheres, while capacity often refers to the volume of a container, such as how much liquid it can hold.

We often come across problems that involve measuring the “surface area” of different objects, whether it is for your science projects, construction tasks or everyday life situations.

Surface Area

“Surface area” refers to the total area of the outer surface of a 3-dimensional object, such as a cube, cylinder or sphere. It is the sum of the areas of all the faces or curved surfaces of the object. For example, the surface area of a cuboid is the sum of the areas of its six faces.

Figure 7.1: Faces, vertices and edges of a cuboid Calculating the Surface Area of given 3D shapes The images below illustrate the surface areas of some solid objects (3D shapes).

Figure 7.2: Cylinder and its net

Figure 7.3: Rectangular prism and its net

Figure 7.4: Rectangular prism and its net

Activity 7.1: Finding the surface areas of solids Materials needed: Toilet roll, Milo tins, milk tins, match boxes, chalk boxes, marker boxes, football and any other geometric object.

Step 1: Sort out the materials into cylinders, spheres, cubes and cuboids.

Step 2: Tabulate these objects with columns headed: Name of object, Net and Total surface area.

Step 3: Discuss with your classmates and establish the various nets and surface areas of the solids. Show them to your teacher for validation.

Hint: You will realise that when you take any cylinder or box and dismantle it, you obtain the net. The surface area of the cylinder or box will be the area of the net or the sum of the areas of all the faces. See figures 7.2, 7.3 and 7.4

Activity 7.2: Finding the surface area of a square pyramid Materials needed:

Paper, ruler and a square pyramid.

Step 1: Identify the number of faces. In the above pyramid, the base is a square with side length 5 cm and each of the four walls is a triangle with base 5 cm and height 8 cm.

Step 2: Find the area of one triangle:

Area of each triangular side wall = 1/2 × base × height = 1/2 × 5 × 8 = 20 cm²Step 3: Area of four triangles = 4 × 20 = 80cm²Step 4: Find the area of the square base:

Area of the square base = 5 × 5 = 25 cm²Step 5: Surface area of the pyramid = Sum of areas of all 5 faces = (25 + 80) cm² Therefore, the surface area of the above pyramid = 105 cm²

Example 7.1

Find the surface area of the triangular based pyramid given below.

Solution

Surface area of the pyramid = Sum of areas of all 4 faces The base is an equilateral triangle with side length 4 cm and each wall is a triangle with base 4 cm and height 6 cm.

Let us find the area of each face.

Area of the base = 1/2 × 2 √3 × 4 = 4√_ 3 cm² (Note that the height of the base triangle is found through Pythagoras’s theorem:

h²+ 2²= 4²∴ h = √12 = 2 √3) Area of each side wall = 1/2 × 4 × 6 = 12 cm² Area of all 3 side walls = 3 × 12 = 36 cm² Surface area of the pyramid = (4√_ 3 + 36) cm² Surface area of the above pyramid = 4(√_ 3 + 9) cm² Real-Life Importance of Surface Area Calculation Calculating surface area is important in various real-life situations, like packaging, manufacturing/construction/architecture.

Packaging To wrap an object, you need to calculate the number of materials needed.

Figure 7.5: Wrapping an object by using the concept of surface area You may also click on the link for a video on the importance of packaging: https://youtu.be/SJGpKnI-784 .

Manufacturing/Construction/Architecture Knowing the surface area helps to establish the cost of painting and designing for functionality and aesthetics (beauty).

Figure 7.6: Surface area in Construction and design

Figure 7.7: Surface Area in architecture The above examples show how calculating surface area is important in everyday tasks, like determining the amount of paint needed for a room. Knowing the surface area helps avoid wasting resources and ensures the job is done efficiently, as well as helps in estimating costs. In industries such as construction, packaging and manufacturing, calculating surface area is critical for tasks such as material estimation, designing and cost planning.

Example 7.2

A painter needs to paint the walls and ceiling of a rectangular room. The room has a length of 8 metres, a width of 6 metres and a height of 4 metres. The door area is 2 metres by 1.5 metres, and there are two windows, each measuring 1.5 metres by 1.5 metres. The painter needs to calculate the surface area of the walls and ceiling to determine how much paint to buy.

How much area does the painter need to cover with paint?

Solution

The room has four walls:

Two walls have dimensions of 8 metres (length) by 4 metres (height).

Two walls have dimensions of 6 metre (width) by 4 metre (height).

Surface area of two 8-metre-long walls:

Area = 2 (8 × 4) = 2 × 32 = 64 Surface area of two 6-metre-wide walls:

Area = 2 (6 × 4) = 2 × 24 = 48 The ceiling has an area of:

Area = 8 × 6 = 48 Door area = 2 × 1.5 = 3 Each window has an area of 1.5 × 1.5 = 2.25.

Total area of two windows = 2 × 2.25 = 4.5.

Total area of walls and ceiling:

64 + 48 + 48 = 160 Areas of the door and windows = (3 + 4.5) = 7.5 The painter needs to cover (160 – 7.5) m²= 152.5m².

Activity 7.3: Applications of surface area in painting your classroom Your classroom paint has faded and needs to be repainted. Your class prefect has drawn the attention of the form teacher who asked for an estimate of the amount of paint required to paint the room. You then contacted a painter who requested the surface area to be covered excluding doors and windows.

Calculate the surface area that needs to be covered with paint.

Activity 7.4: Applications of surface area in a home You are tasked with designing a small swimming pool for a family home. The pool has a rectangular shape with a length of 10 metres, a width of 5 metres and a depth of 2 metres.

Calculate the surface area to determine the quantity of tiles required to cover the pool’s walls and floor.

In small groups, discuss why calculating the surface area is important in real-life situations like construction or design, especially when considering material costs.

Units of Measurement

Measurement is fundamental in science, engineering and everyday life. Different countries and industries use various systems of measurement. Two of the most widely known systems are the “Imperial system” and the “International System of Units (SI)”, also known as the “Metric System”.

1. Imperial System: This is commonly used in countries like the United States and the United Kingdom, although the metric system is becoming more common in the UK. It includes units such as pounds (lbs), feet (ft), inches (in) and gallons.

2. SI (Metric) System: This is the globally adopted system used by almost all countries including Ghana. It includes units such as kilograms (kg), metres (m), and litres (l).

The ability to convert between these two systems is important, especially in scientific, technological and international contexts.

Conversion between Imperial and SI (Metric) Units We will often come across both the Imperial (British) system and SI system of measurement. This guide will help us understand how to convert between these two systems, focusing on commonly used units of measurement for length, area, mass and volume.

Study the table below:

Table 7.1: Metric and Imperial units Metric Units Imperial Units 1 centimetre 0.394 inches 1 metre 3.281 feet = 1.093 yards 1 kilometre 0.621 miles 1 gramme 0.035 ounces 1 kilogram 2.205 pounds 1 millilitre 0.034 fluid ounces 1 litre 1.057 quart = 0.264 gallons 25.4 millimetres 1 inch 0.3048 metres 1 foot = 12 inches 0.9144 metres 1 yard = 3 feet 1.60934 kilometres 1 mile = 1 760 yards The above table illustrates the relationship between metric units and imperial units.

These are called conversion factors and are used to convert one unit to another.

Example 7.3

Convert 5 m²to cm²Solution Here we are converting within the metric system.

5 m²= 5(1m × 1m) = 5(100cm × 100cm) = 5 ×10 000 = 50 000 cm²Note: To convert square metres to square centimetres, multiply by 10 000.

Example 7.4

Convert 85 000 cm²to m²Solution 85 000 cm²= 85 000/10 000 = 8.5 m²Note: To convert square centimetres to square metres, divide by 10 000.

Example 7.5

Convert 8 ft²to in².

Solution

Here we are converting within the imperial system. Remember that 1 foot = 12 inches.

8 ft²= 8(1feet × 1 feet) = 8(12 × 12) = 8 ×144 = 1 152 in²Note: To convert square feet to square inches, multiply by 144.

Example 7.6

Convert 310 in²to ft²Solution 310 in²= 310/144 = 2.152777 ft²Note To convert square inches to square feet, divide by 144.

Guidelines for Converting Units of measurement:

1. Identify the appropriate unit to convert (either “from” or “to”).

2. Identify the conversion factor.

3. Multiply or divide using the conversion factor.

Table 7.2: Summary of some Key Conversion Formula Quantity Unit (Imperial or SI) Unit (SI or Imperial) Conversion Factor Length 1 inch 2.54cm 1 in = 2.54cm 1 foot 0.3048m 1 ft = 0.3048m 1 yard 0.9144m 1 yd = 0.9144m 1 mile 1.60934km 1 mile = 1.60934km Area 1 m²10 000 cm²1m²= 10 000cm² 1 m²10.7639 ft²1 m²= 10.7639 ft² 1cm²0.155 in²1cm²= 0.155 in² 1 ft²144 in²1 ft²= 144 in² 1 ft²0.092903 m²1 ft²= 0.092903 m² 1 in²6.4516 cm²1 in²= 6.4516 cm² Mass 1 pound 0.453592kg 1 pound = 0.453592kg 1 ton 907.1847kg 1 ton = 907.1847kg 1 kilogram 1 000g 1kg = 1 000grams Volume 1 gallon (US) 3.78541litres 1 gal (US) = 3.7854 l 1 gallon (UK) 4.54609 litres 1 gal (UK) = 4.54609 l 1 litre 1 000 cm³1 litre = 1 000cm³ By understanding and practising these conversions, we can efficiently handle different units of measurements. Understanding and mastering conversions between the Imperial and SI systems is essential, especially for progress in science and mathematics and in general, everyday tasks.

Example 7.7

Convert 3 gallons (UK) to litres.

Solution

1 gallon (UK) = 4.54609 litres 3 gallons (UK) = 3 × 4.54609 litres Therefore, 3 gallons (UK) = 13.63827 litres

Activity 7.5: Converting Imperial to SI and Vice Versa The measurements of a rectangular room are:

length of the room = 20 feet, width = 15 feet, height = 10 feet.

Convert these measurements into the metric system (SI units) to calculate the volume of the room in cubic metres.

Steps to do this:

1. Convert the room’s length, width and height from feet to metres, remember, 1ft = 0.3048m.

2. After converting the dimensions, calculate the volume of the room in cubic metres.

3. In small groups, discuss why it is important to understand how to convert between Imperial and SI units when working on real-world problems.

Nets of 3D objects (prisms, cones, pyramids, spheres, etc.)

We will now explore nets of 3D shapes. Below we can see nets of cuboids. Imagine taking the 3D shape and dismantling it and spreading it out to form the net.

Net of a hexagonal prism

Figure 7.8: Nets of 3D shapes Click on the link for a video on nets of solids. Then carry out the

activity below: https://youtu.be/s7GrS0b3FRw.

Activity 7.6: Nets of 3D objects

Figure 7.9: Nets of 3D shapes.

Activity 7.7: Finding nets of 3D shapes Working in small groups, draw the nets of the 3D shapes in figure 7.10 below.

Compare your results with other groups and discuss any differences.

Figure 7.10: Nets of 3D objects

All three-dimensional (3-D) objects take up a certain amount of space, this is known as their volume. Conversely, capacity refers to the maximum amount of substance or material that a 3D object can hold. Therefore, capacity is the container’s volume. Volume is measured in cubic units, whereas, capacity is often measured in litres, millilitres or gallons.

Figure 7.11: Measurement of Volume and Capacity

Click on the link for more information on volume and capacity: https:// youtu.be/xSbsbz7Ovb4, or search the internet for more information on volume and capacity.

Surface Area measures the area occupied by the outer layer of a solid. Or we can think of it as the area of all the shapes/planes that make up the outside of the figures (solids). In contrast, volume measures the carrying capacity of a figure/ shape or the space enclosed within the formation.

Table 7.3: Differences between volume and capacity Volume Capacity Volume indicates the total amount of space covered by an object in three-dimensional space.

Capacity refers to the ability of something to hold, absorb or receive other substances.

Common units of measurement = cm³, m³ Common units of measurement = litres, gallons Both solid and hollow objects have volume.

Only hollow objects have capacity.

Example – Cube, Cuboid, Cone and Cylinder

Example – Only hollow shapes such as a hollow cone or hollow hemisphere Calculating volumes and capacities of 3D shapes

Example 7.8

Calculate the volume of the cuboid below.

Solution

Since the base is a rectangle and all the side walls are also rectangles, we have;

Volume of the cuboid = Base Area × Height Area of base = 12 × 4 = 48 cm² Height of the cuboid = 8 cm.

Volume of cuboid = 48 × 8 = 384 cm³

Example 7.9

Find the volume of pyramid given below.

Solution

Volume of the pyramid = 1/3 × Base Area × Height Area of the base = 8 × 8 = 64 cm².

Height of the pyramid = 9 cm Volume of the pyramid = 1/3 × 64 × 9 = 192 cm³Example 7.10 Study the composite shape below.

The shape is a cone attached to the top of a cylinder.

The base of the cylinder has radius 4 mm, the height of the cylinder portion is 3 mm and the height of the cone is 5.5 mm.

Calculate the volume of the whole shape to the nearest whole number.

Solution

To work out the volume of the shape, we need to work out the two volumes separately.

Volume of cylinder = π × 4²× 3 = 48π Volume of cone = 1/3π × 4²× 5.5 = 88/3 π Then, the volume of the shape is the sum of these two answers:

Volume of whole shape =48π + 88/3 π = (48 + 88/3 )π = 232/3 π = 242.9498mm³ = 243mm³to the nearest whole number.

Example 7.11 (CHALLENGE) A solid sphere and a solid hemisphere have equal total surface area. Prove that the ratio of their volumes is 3√_ 3 : 4.

Solution

Let the radius of the sphere = r₁ Let the radius of the hemisphere = r₂ ∴ Total surface area of sphere = 4πr₁ ²∴ Total surface area of hemisphere = 3πr₂ ² 4πr₁ ²= 3πr₂ ² r₁ ²= (3/4)r₂ ² r₁= (√3/2 )r₂ Volume of sphere = (4/3) πr₁ ³ Volume of hemisphere = (2/3) πr₂ ³(4/3) πr₁ ³: (2/3) πr₂ ³4/3(√3/2 )³r₂ ³: 2__ 3r₂ ³12 √3/24 r₂ ³: 2__ 3r₂ ³36 √3 : 48 3√_ 3 : 4 as required.

Volume and capacity have significant real-life importance across various fields and everyday activities. Here are a few key areas where these concepts are commonly applied:

1. Cooking and Food Preparation: When measuring ingredients like liquids (for example, water, milk, oil) or solids (for example, flour, sugar), volume measurements are crucial. For example, recipes may call for “1 cup of flour” or “500 millilitres of water”. Containers, such as pots, cups or measuring spoons, are designed to hold a certain capacity. The capacity of these items determines how much they can hold, affecting how much food can be prepared at once.

Figure 7.11: Measurement of capacity in food preparation

2. Healthcare: Medical professionals use volume measurements to administer medications in liquid form (for example, syringes often measure in millilitres) or to monitor the volume of fluids in the body, such as blood volume in patients or patients are prescribed a teaspoon of liquid medication.

Figure 7.12: Measurement of volume and capacity in healthcare The capacity of medical devices, such as IV bags, oxygen tanks and dialysis machines are crucial in delivering appropriate amounts of fluids or gases to patients.

3. Transportation and Shipping: The volume of goods being transported;

whether in shipping containers, trucks, or ships, affects how much cargo can be moved at once. Volume calculations ensure that space is used efficiently.

Figure 7.13: Measurement of Volume and Capacity in transportation Vehicles, such as trucks, buses or airplanes, have capacity limits for passengers and cargo, which must be adhered to for safety and efficiency.

For example, a shipping container’s capacity dictates how much freight can be packed.

4. Architecture and Construction: Volume calculations are used in construction for determining the amount of material needed for projects (for example, how much concrete is required to fill foundations).

Figure 7.14: Measurement of Volume and Capacity in Architecture In buildings, the capacity of rooms, elevators and facilities must be considered to ensure that they can accommodate the expected number of people or goods. For example, an auditorium’s seating capacity or the capacity of a water tank for a building’s plumbing system.

5. Environmental and Energy Systems: Volume is used to measure environmental factors like air and water pollution, where the amount of substance per unit volume (for example, mg per litre) can affect health and safety standards.

Figure 7.15: Measurement of Volume and Capacity in environment and energy systems Energy storage systems, such as batteries or fuel tanks, rely on capacity measurements to ensure they can store enough energy for specific tasks, such as powering an electric vehicle or providing backup power during outages.

6. Retail and Manufacturing: In retail, products like liquids (for example, beverages or cleaning products) are sold based on volume (for example, litres, gallons).

Figure 7.16: Volume and Capacity in retail and manufacturing The capacity of storage systems, such as warehouse shelves, refrigerators and production lines, determine how much can be stored or produced, affecting inventory management and supply chain logistics.

7. Personal and Daily Life: Everyday tasks like filling up a gas tank, pouring a drink or using a shower all involve volume calculations. Knowing the volume of a container (like a water bottle) is important for determining how much it holds.

Figure 7.17: Measurement of Volume and Capacity in daily life Understanding the capacity of a bag, suitcase or even a parking space helps in making decisions about how much can fit or how much space is available.

Therefore, we can see that volume and capacity are essential concepts that impact both the efficiency and effectiveness of everyday tasks, from managing household

activities to ensuring safety and functionality in industries like healthcare, transportation, and construction.

Common Metric Units for Volume

1. Cubic Metres (m³): The cubic metre is the primary metric unit for measuring large volumes in the metric system. It is often used for measuring the volume of rooms, storage areas, or large containers.

2. Cubic Millimetre (mm³): The cubic millimetre is a smaller metric unit typically used in scientific fields, especially in chemistry and medicine, to measure volumes of liquids and solids in lab experiments.

3. Litres (l): The litre is a commonly used metric unit for measuring liquids, particularly in everyday settings such as beverages, fuel and household items. For example, liquids like milk, water and soft drinks are typically measured in litres.

4. Millilitres (ml): The millilitre, a smaller metric unit than the litre and it is frequently used in cooking, medicine and scientific research for measuring small volumes, such as ingredients in recipes or medication dosages.

Common Imperial Units for Volume

1. Cubic Feet (ft³): The cubic foot is used for measuring larger volumes in areas like storage, shipping, and construction. It is commonly used to describe the volume of appliances, such as refrigerators and freezers.

2. Cubic Inches (in³): The cubic inch is a smaller imperial unit often used in engineering and automotive industries to measure things like engine displacement and small components.

3. Gallons: Gallons are widely used in the United States for measuring larger volumes of liquids, including fuel, milk and water. It is important to note that the U.S. gallon differs from the imperial gallon used in the UK.

4. Quarts: A quart represents a quarter of a gallon and is often used for measuring smaller quantities of liquid, such as ingredients in recipes and products like motor oil.

5. Pints: A pint is half of a quart and is commonly used to measure smaller liquid volumes, such as drinks (for example, beer, milk) or food items like ice cream.

6. Fluid Ounces: Fluid ounces are the smallest common imperial unit for liquid volume, used for very small quantities such as drink servings, cooking ingredients and doses of medicine.

Conversion between SI and Imperial units When converting between SI (metric) units and imperial units of volume and capacity, the following formulae and conversion factors are commonly used:

SI to Imperial Units

1 Cubic Metre (m³) = 35.3147 Cubic Feet (ft³)

Example 7.12

You have a container with a volume of 2 cubic metres, Convert this to cubic feet.

Solution

2 cubic metres = 2m³× 35.3147 ft³/m³= 70.6294 ft³1 Litre (l) = 0.21997 British gallons

Example 7.13

You have a 75-litre container of water. Convert this to UK gallons.

Solution

75 litres = 75 l × 0.21997 gal/l = 16.49775 gal

Note: 1 litre = 0.264172 US gallons Imperial to SI Units 1 Cubic Foot (ft³) = 0.0283168 Cubic Metres (m³)

Example 7.14

A storage polytank has a volume of 34 cubic feet. Convert this to cubic metres.

Solution

34 cubic feet = 34 ft³× 0.0283168 m³/ft³= 0.9627712 m³1 British (Imperial) gallon is equal to 4.54609 litres.

Example 7.15

You have a 50-gallon (UK) storage tank. Convert this to litres.

Solution

50-gallon (UK) = 50 gal × 4.54609 1 l/gal = 227.3045 l Calculating Surface area and volume of 3D shapes Study the 3D shapes in Figure 7.18 and make sure you can understand how the formulas are derived.

Figure 7.18: Calculating surface area and volume of 3D shapes You may also click on the link: https://youtu.be/eBAq_caikJ4 for further clarifications on surface area and volumes of prisms.

Example 7.16

A rectangular roof measures 12 metre in length and 8 metre in width. You need to cover the entire roof with roofing tiles. Each roofing tile covers an area of 0.5 square metre. How many tiles are required to cover the entire roof?

Solution

Area of the roof = length × width = 12 metre × 8 metre = 96 square metres.

Number of tiles = Total area ÷ Area covered by each tile = 96 square metre ÷ 0.5 square metre/tile = 192 tiles Therefore, 192 tiles are required to cover the roof.

Volumes of Pyramids

There are different types of pyramids. The name of pyramids is derived from its base shape. Different types of pyramids are shown in the table below. Their volume is always calculated by this formula:

Volume of Pyramid = 1/3 × Area of Base × Vertical Height

Table 7.4: Volumes of pyramids

Let us look at the following example.

Example 7.17

You are designing a new water reservoir in a community garden, and the reservoir will be in the shape of a frustum of a cone (a cone with its top sliced off). The reservoir has the following dimensions: The diameter of the bottom is 10 metres and the diameter of the top is 6 metres. The height of the frustum is 8 metres. The community plans to use this reservoir to collect rainwater for irrigation.

1. Calculate the volume of the frustum of the cone to determine how much water it can hold in cubic metre.

2. Convert the volume into litres, knowing that 1 cubic metre equals 1 000 litres and explain how this helps in estimating the amount of water the reservoir can store.

Solution

1. V_(fc) = 1/3 πH(r²+ r r¹+ r₁ ²) Where V_(fc) = Volume of frustum of cone H = Height of frustum = 8m r = radius of the bottom = 10/2 = 5m r¹= radius of the top = 6/2 = 3m Now, substituting into the formular, we have;

V_(fc) = 1_ 3 π8(5²+ 5 × 3 + 3²) V_(fc) = 1_ 3 π8(25 + 15 + 9) V_(fc) = 1/3 π8(49) = 392/3 π V_(fc) = 410.5024 m³2. If 1m³= 1000litres Then, 410.5024 m³= 410.5024 m³× 1000 litres/m³= 410 502.40 litres The reservoir can hold approximately 410 502 litres of water, which helps the community estimate how much rainwater it can collect for irrigation.

Since a typical watering system for a garden may require a few hundred litres of water per session, this volume is quite substantial and would allow the garden to be watered over many sessions before needing to refill.

Example 7.18

Amina is interested in knowing how much her family spends on water for showers.

Water costs GH¢60.00 for 1,000 gallons. Her family averages 6 showers per day. The average length of a shower is 15 minutes. She places a bucket in her shower and turns on the water. After one minute, the bucket has 3.5 gallons of water.

About how much money does her family spend on water for showers in a 30- day month?

Solution

Step 1: Total time of showers per day 6 showers × 15 minutes = 90 minutes

Step 2: Gallons used per day 90 minutes × 3.5 gallons / minute = 315 gallons / day

Step 3: Gallons used per month 315 × 30 = 9450 gallons / month

Step 4: Cost of 9450 gallons Since 1000 gallons = GH¢60.00, Then 9450/1000 × GH¢60.00 = 9.45 × 60 = GH¢567.00 Answer The family spends approximately GH¢567.00 on water for showers in a 30- day month.

Example 7.19

Ama loves playing with building blocks. She built a structure with 10 cubic blocks.

If the edge of each cube is 3.5 inches, what is the volume of her structure?

Solution

The volume of a cube = Length³= 3.5³= 42.875in³ There are 10 cubes in her structure.

Volume of the structure = 10 × Volume of one cube = 10 × 42.875 in³= 428.75 in³Therefore, the volume of her structure is 428.75 in³.

Example 7.20

Kofi has a glass that is cylindrical in shape. The height of the glass is 20 units and the radius of the base is 5 units.

What is the capacity of the glass?

Solution

Height of the glass = 20 units and the radius = 5 units.

Volume of a cylinder = πr²h cubic units.

Volume of the glass, V = πr²h = π × (5²) × 20 = π × 500 = 1570.8 cubic units.

Therefore, the glass’s capacity is approximately 1571 cubic units.

Find the surface area of each of the following solid shapes in cm²,correct to 1 decimal place.

1. 2. 3.

4. Convert 60 miles to kilometres.

5. The net of a triangular prism is given below.

Find the surface area (the triangles are equilateral triangles).

6. What is the total surface area of a cylinder whose radius is 4.5 units and height is 8 units.

7. You are building a raised rectangular garden bed that is 3 metres long, 2 metre wide and 0.5 metre deep. How much soil (in cubic metres) is required to fill the garden bed?

8. You are designing a new water reservoir in a community garden and the reservoir will be in the shape of a frustum of a cone (a cone with its top sliced off). The reservoir has the following dimensions: The diameter of the bottom is 12 metres, and the diameter of the top is 8 metres. The height of the frustum is 10 metres. The community plans to use this reservoir to collect rainwater for irrigation.

a. Calculate the volume of the frustum of the cone to determine how much water it can hold in cubic metre.

b. Convert the volume into litres, knowing that 1 cubic metre equals 1 000 litres, and explain how this helps in estimating the amount of water the reservoir can store.

9. Find the surface area of the cuboid below.

10. The volumes of two cones of same base radius are 3600 cm³and 5040 cm³.

Find the ratio of their heights.

11. Find the volume of triangular prism given below.

12. The radius of the Earth is about 4000 miles. The radius of the Sun is about 400 000 miles.

How many times bigger than the Earth is the Sun?

In this section, we will explore two key areas: vector operations and trigonometry.

The focus will be on the addition, subtraction and scalar multiplication of vectors. We will uncover the properties that govern these operations, such as the commutative, associative and distributive laws and how these concepts apply to real-life situations. We will then review basic trigonometric ratios and extend our knowledge to inverse trigonometric functions which enables the calculation of angles when the ratio values are known. Finally, we will explore the real-world applications of trigonometry in various fields of mathematics

KEY IDEAS

• Directed line segment is a segment with an initial point and a terminal point and it is called a vector.

• Position vector is a vector that describes the position of a point on the Cartesian plane relative to the origin resultant vector.

• Scalar quantity is a quantity that has magnitude only. For example, mass, length, time, temperature, density, speed are scalar quantities.

• Trigonometry is the branch of mathematics which deals with the measurements of angles and lengths of sides of triangles and their applications.

• Vector quantity has both magnitude and direction. For examples, force, velocity, acceleration and momentum are vector quantities.

The concept of vectors Vectors are mathematical entities with magnitude and direction, used to represent quantities in space. They can be added, scaled and multiplied using operations like vector addition, scalar multiplication, dot product and cross product. These operations enable calculations of angles, lengths and orientations between vectors, providing a powerful tool for problem-solving.

Real–life application of vectors

1. Walking to School: Imagine your teacher asks you to fetch a book from another classroom. If they say, “Walk 30 metres straight ahead and then turn left for 20 metres,” they are giving you a vector. The distance (30 metres and 20 metres) is the magnitude and the directions (straight ahead and left) define the path.

2. Playing Football: When you pass a ball to a teammate, the ball travels in a specific direction with a certain force. For example, kicking the ball 10 metres towards the goalpost represents a vector because it combines the distance (10 metres) with the direction (towards the goalpost).

Vector as a Visual Representation

Vectors can be visually represented as arrows on a diagram or graph. The length of the arrow shows the magnitude (size) of the vector, and the direction of the arrow shows the direction of the vector. Let us explore this with some examples:

1. Walking to the Market

Imagine you are walking to the market:

a Start from your house and walk 3 kilometres east. Then turn and walk 4 kilometres north to reach the market.

b Draw an arrow pointing east for the first part (3 km) and another arrow pointing north for the second part (4 km).

c To find the direct distance to the market, draw a single arrow from the starting point to the market. This is the resultant vector.

2. Flying a Paper Plane

When you throw a paper plane:

a If you throw it straight, the arrow representing the vector points in the direction of the throw.

b The longer the arrow, the harder the throw (the greater the magnitude).

c For example, a throw 5 metres away in a north-east direction can be drawn as an arrow starting from you and pointing and ending at the plane’s landing spot.

Definition of terms In our everyday life, we encounter two main types of quantities: scalar and vector:

A scalar is a quantity that has only magnitude, without any specific direction.

Examples of scalars include:

1. Mass

2. Length

3. Time

4. Temperature

5. Density

6. Speed These quantities only tell us how much there is, without any reference to direction.

A vector is a directed line segment, which has both an initial point and a terminal point. It is a quantity that has both magnitude (size) and direction.

Examples of vectors include:

1. Force

2. Velocity

3. Acceleration

4. Momentum These quantities not only tell us how much there is but also in which direction.

Position Vector

A position vector is a vector that represents the position of a point or object relative to a reference point, usually the origin of a coordinate system. It indicates both the magnitude (distance) and the direction from the origin to the point.

In simpler terms, a position vector tells you where something is located in space, relative to a defined starting point (usually the origin (0,0) in a 2D plane.

If M (x, y ) is any point on the Cartesian plane and O is the origin, then the point M relative to O is a position vector ⟶ OM. That is ⟶ OM = ( x y)

Figure 9.1: Position vector ⟶ OM For example, the point M (4, 5) on the Cartesian plane will be a position vector ⟶ OM = (4 5)

Figure 9.2: Position vector ⟶ OM = (4 5) Expressing two given points as a Vector Suppose the two points are A( x₁, y₁) and B( x₂, y₂) The vector from A to B is given by:

⟶ AB = ( x₂ − x₁, y₂ − y₁) For example: If A (2, 3) and B (5,7) then:

⟶ AB = (5 − 2, 7 − 3) = (3, 4) This vector, ⟶ AB = (3, 4), indicates the direction and magnitude from point A to point B.

Figure 9.3: A vector from A to B When we combine two or more vectors, like forces acting on an object or the velocity of moving objects, we are finding their total effect. This total effect is called the resultant vector. It shows the overall result when the vectors are added together. For example, if you walk 3 kilometres north and then 4 kilometres east, the resultant vector shows your final position from where you started. We can use simple maths to find the size and direction of the resultant vector, which we will learn about later.

Understanding Resultant Vectors

A resultant vector is the combined effect of two or more vectors. It shows the overall direction and size of the combined vectors. Imagine several forces acting on an object; the resultant vector represents the single force that would have the same impact.

Visualising Resultant Vectors

Activity 9.1: Using the Head-to-Tail Method to visualise vectors Working individually, or in pairs, follow these steps to visualise resultant vectors using the Head to Tail Method:

Step 1: Begin by drawing the first vector to scale, with its direction and magnitude accurately represented.

Step 2: Position the tail of the second vector at the head (arrow tip) of the first vector.

Step 3: If there are more vectors, repeat the process by placing the tail of each new vector at the head of the previous one.

Step 4: Draw a straight line from the tail of the first vector to the head of the last vector. This line represents the resultant vector.

Figure 9.4: Resultant vector

Step 5: Use a ruler to measure the length (magnitude) of the resultant vector and a protractor to determine its direction relative to a reference axis.

Key Points to Remember:

• The order in which you add the vectors doesn’t matter. The resultant will always be the same.

• Always use a scale and, where applicable, indicate units for clarity.

Activity 9.2: Using the Parallelogram Method to visualise vectors Working individually, or in pairs, follow these steps to visualise resultant vectors using the Parallelogram Method:

Step 1: Begin by drawing the two vectors to scale, starting from the same origin point. Ensure their direction and magnitude are accurately represented.

Step 2: From the head of each vector, draw lines parallel to the other vector to form a parallelogram. These lines should be drawn to the same scale as the original vectors.

Figure 9.5: Parallelogram law

Step 3: Draw a diagonal line of the parallelogram starting from the origin point where the two vectors meet. This diagonal represents the resultant vector.

Step 4: Use a ruler to measure the length (magnitude) of the resultant vector and a protractor to determine its direction relative to a reference axis.

Calculating Resultant Vectors

If you know the magnitudes and angles of the individual vectors, you can use trigonometry to determine the magnitude and direction of the resultant vector.

Activity 9.3: Using the Head to Tail Method to find the resultant vector In small groups work through the following steps to find the resultant vector.

Two forces act on an object. One force has a magnitude of 12 N and the other has a magnitude of 16N. The angle between the two forces is 40° .

Find the magnitude and direction of the resultant force.

Step 1: Draw the Vectors:

Draw the First Vector: Represent the 12 N force as a vector. Let’s call this vector A Draw the Second Vector: Represent the 16 N force as another vector. Let’s call this vector B Align the Vectors: Place the tail of vector B at the head of vector A, ensuring that the angle between them is 40°. Imagine sliding the tail of vector B to the tip of vector A without changing the angle between them. Note, the interior angle between them is 140°, compared with the 40° on the exterior.

Figure 9.6: Resultant vector

Step 2: Determine the Resultant Vector:

Connect the Tail of A to the Head of B: Draw a line from the tail of vector A to the head of vector B. This line represents the resultant force vector, let’s call it R.

Step 3: Calculate the Magnitude of the Resultant Force To calculate the magnitude of the resultant force, you can use the cosine rule since you know the magnitudes of the two forces and the angle between them.

Cosine Law Formula:

R²= A²+ B²− 2ABcos(θ) Where:

R is the magnitude of the resultant force.

A is the magnitude of force A (12 N).

B is the magnitude of force B (16 N).

θ is the angle between the two forces (140°).

Calculation:

R²= 12²+ 16²− 2 × 12 × 16cos(140°) R²= 144 + 256 − 384cos(140°) R²= 400 − 384 × (−)0.766… R²= 400 + 294.161… R²= 694. 161… R = √694.161… ≈ 26.3N 12N 16N 140° 40° 0°

Figure 9.7: Resultant vector

Step 4: Formula for Direction (Angle)

To find the angle θ between the resultant force and A and B, use the Sine Rule or a trigonometric formula:

sinθ_ 12 = sin140/26.347 sinθ = 12sin140/26.347 θ = sin⁻¹(12sin140_ 26.347 ) = 17.0° Magnitude of Resultant Force: Approximately 26.3 N Direction of Resultant Force: Approximately 17° above vector B.

Activity 9.4: Using the Parallelogram Method to find the resultant vector In your small groups work out Activity 9.3 using the parallelogram method.

As you will see, they are very similar in their method to give the same solutions.

Example 9.1

There are two vectors:

Vector A: Magnitude = 5 units, Direction = 30° from the vertical Vector B: Magnitude = 10 units, Direction = 30° below the horizontal Find the magnitude of the resultant vector.

Solution

To find the resultant vector, you would use the cosine rule to calculate its magnitude.

OA = (5cos60° 5sin60°) = ( 2.5 2.5 √3) OB = ( 10cos30° − 10sin30°) = (5 √3 − 5 ) OA + OB = ( 2.5 2.5 √3) + (5 √3 − 5 ) = ( 11.16 − 0.67) Magnitude = √_______________ (11.16)²+ (− 0.67)²= √___________ 124.55 + 0.45 ≈ √125 = 11.18 units

Example 9.2

Find the sum of the two given vectors m and n .

Solution

Draw the vector m .

Draw the ‘tail’ of vector n , joined to the ‘nose’ of vector m .

The vector m + n is from the ‘tail’ of m to the ‘nose’ of n .

Geometric representation A vector v = (x, y) can be represented geometrically as an arrow:

1. The tail of the arrow is at the origin (0, 0).

2. The head of the arrow is at the point (x, y).

3. The length (or magnitude) of the arrow represents the size of the vector.

4. The direction of the arrow represents the direction of the vector.

Algebraic Representation

A vector in two-space can be represented algebraically as:

v = (x, y) or v = x i + y j where:

x and y are the horizontal and vertical components of the vector, respectively and i and j are the unit vectors in the x and y directions, respectively.

The relationship between geometric and algebraic representations:

1. x is the change in x-coordinate from tail to head

2. y is the change in y-coordinate from tail to head

3. Zero vector: (0, 0)

4. Unit vectors: î = (1, 0) a nd ĵ = (0, 1)

5. General form: a = xî + y ĵ

Activity 9.5: Using graph paper to illustrate the connection between geometric and algebraic representations Working individually, or in pairs, follow the steps below to illustrate the connection between geometric and algebraic representations.

Step 1: On graph paper mark on the x and y axes.

Step 2: Select two vectors, A and B, with their components (x, y). For

example, A = (3, 4) and B = (2, 1).

Step 3: Plot Vector A by starting at the origin (0, 0) and move horizontally 3 units (x-component) and then vertically 4 units (y-component). Mark the endpoint as the head of Vector A.

Step 4: Plot Vector B by starting at the origin (0, 0) and move horizontally 2 units (x-component) and then vertically 1 unit (y-component). Mark the endpoint as the head of Vector B.

Step 5: Label Vector A and Vector B with their corresponding components (x, y).

Step 6: To illustrate vector addition, place the tail of Vector B at the head of Vector A. Draw the resultant vector from the tail of Vector A to the head of Vector B.

Step 7: Verify that the geometric representation of the vectors matches their algebraic representation.

For example, Vector A = 3i + 4j, Vector B = 2i + 1j and Vector A + B = 5i + 5j Vector addition and subtraction Vector Addition Geometrically:

1. Head-to-tail method: Align the tail of the second vector with the head of the first vector. The resulting vector is represented by an arrow drawn from the tail of the first vector to the head of the second.

2. Parallelogram method: Construct a parallelogram where the two vectors form adjacent sides. The diagonal of the parallelogram, starting from the same point as the vectors, represents their sum.

Note that this is the same as the Resultant Vector investigated above.

Algebraically:

For vectors m = (x₁, y ₁) and n = (x₂, y ₂):

m + n = (x₁+ x₂, y ₁+ y ₂)

Example 9.3

Given that, ⎯⇀AB = (2

3) and ⎯⇀BC = ( 2 − 2), find the sum of the vectors

Solution

The sum of the vectors ⎯⇀AB and ⎯⇀BC is the same as the vector ⎯⇀AC.

That is ⎯⇀AB + ⎯⇀BC = ⎯⇀AC.

Therefore, we add the corresponding components of the vectors as:

⎯⇀AB + ⎯⇀BC = (2

3) + ( 2 − 2) = (4 1) ⎯⇀AC = (4 1) Vector Subtraction Subtraction as Vector Addition Subtracting a vector B is equivalent to adding its negative, −B, which is the same vector pointing in the opposite direction: A − B = A+ (−B) Steps for Geometric Representation:

1. Plot vector A.

2. Reverse the direction of vector B to obtain –B.

3. Use the head-to-tail method to add A and −B.

4. Draw the resultant vector from the tail of A to the head of −B

Example 9.4

m = (4

3) and n = (2 1) Find m – n

Solution

1. Plot m and n on a Cartesian plane.

2. Reverse the direction of n to get −n=(− 2 − 1).

3. Add m and –n m – n =(4

3) +(− 2 − 1) = (2 2) Or think of it as:

m – n =(4

3) − (2

1) = (2 2) Key concepts:

1. Vector addition is commutative: a + b = b + a

2. Vector addition is associative: (a + b ) + c = a + (b + c)

3. The zero vector is the additive identity

4. The negative of a vector: −a = (−x, −y ) Scalar multiplication and its effect on vector magnitude and direction Scalar multiplication in vector mathematics involves multiplying a vector by a scalar (a real number). This operation affects both the magnitude and sometimes the direction of the vector:

Effect on Magnitude:

The magnitude of a vector scales by the absolute value of the scalar.

1. If v is a vector and k is a scalar, then the magnitude of kv is given by:

a ∣ kv ∣ = ∣ k ∣ ⋅ ∣ v ∣

2. If ∣ k ∣ > 1 , the vector’s magnitude increases.

3. If 0 < ∣ k ∣ < 1 , the vector’s magnitude decreases.

4. If k = 0 , the vector becomes the zero vector, which has zero magnitude.

Effect on Direction

The direction of the vector changes based on the sign of the scalar:

1. If k > 0 , the direction of the vector remains the same.

2. If k < 0 , the direction of the vector reverses (points in the opposite direction).

Unit Vector

A unit vector is a vector with a magnitude of 1. It is primarily used to indicate the direction of a vector without considering its magnitude. Unit vectors are widely used in mathematics, physics, and engineering.

To find the unit vector ˆu in the direction of a given vector v , divide the vector by its magnitude:

ˆu = v__ |v|

Example 9.5

v = (3 4) Find its unit vector

Solution

Step 1: Compute the magnitude:

|v| = √3²+ 4²= 5

Step 2: Divide each component by the magnitude:

ˆˆu = 1/5(3

4) = ( 3/5 4/5) ∴ unit vector ˆv = ( 3/5 4/5) Properties of scalar vectors

1. Distributive property: (a + b) = ka + kb

2. Associative property: (kl ) = (l a), where k and l are scalars

3. Vector decomposition: a = xa î + y a ĵ

Table 9.1: Property of Vector Addition

Property of Vector

Addition Explanation

Existence of identity For any vector v, v + 0 = v Here, 0 vector is the additive identity Existence of inverse For any vector v, v + - v = 0 and thus, an additive inverse exists for every vector.

Commutativity Addition is commutative; for any two arbitrary vectors c and d, c + d = d + c Associativity Addition is associative; for any three arbitrary vectors i, j and k, i + j + k = i + k + j i.e., the order of addition does not matter.

Example 9.6

If x = (3

4) and y = ( 2 − 1), find x + y

Solution

x + y = (3

4) + ( 2 − 1), x + y = (3 4 + 2 − 1) x + y = (5 3)

Example 9.7

Given vectors m = (3

4) and w = ( 2 − 1), find m + w

1. algebraically

2. geometrically.

Solution

1. Algebraically:

m + w = (3

4) + ( 2 − 1) = (3 + 2 4 − 1) = (5 3)

2. Geometrically:

• Draw vector m from the origin to point (3 4). From the head of m , draw on the vector w ( 2 − 1)

• The resultant is vector n which is (5 3) b

Example 9.8

Subtract vector p = (3

2) from vector q = (5 7)

Solution

q − p = (5

7) − (3

2) = (5 − 3 7 − 2) = (2 5)

Example 9.9

Multiply vector v = (2

3) by scalar k = 2.

Solution

k . v = 2(2

3) = (2 × 2 2 × 3) = (4 6) Real-life uses of vectors

1. Vectors can be used in finding the direction in which the force is applied to move an object.

2. The concept of vectors aids in understanding how gravity uses a force of attraction on an object to work.

3. Vectors can be used in obtaining the motion of a body which is confined to a plane.

4. Vectors help in defining the force applied on a body simultaneously in the three dimensions.

5. In the field of Engineering, for a structure not to collapse, vectors are used where the force is much stronger than the structure will sustain.

6. Vectors are used in various oscillators.

Trigonometric ratios Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles, particularly triangles with right angles (90-degree angles). The term “trigonometry” comes from the Greek words “Trigonon” meaning “triangle” and “metry” meaning “measure”. Trigonometry involves the study of Core Concepts such as Angles, Triang les and Trigonometric functions.

Right-Angled Triangle

Activity 9.6: Construction of a right -angled triangle using a ruler and a compass

Step 1: Draw a line segment from P to A of any length. Mark its midpoint with the letter C.

Step 2: Place the compass point on one end of the base say A and draw an arc.

Step 3: Move the compass point to the other end of the base, P, and draw another arc, with the same radius, intersecting the first arc.

Step 4: Draw a line from the intersection point to point C on the line PA, creating a right angle (90°).

Step 5: Label the vertices: A (base), B (intersection), and C (right angle).

Figure 9.9: Right -angled triangle Pythagoras’ Theorem Pythagoras’ theorem states the relationship between the lengths of the three sides of a right-angled triangle. It states that a²= b²+ c²as shown below.

Figure 9.10: Right -angled triangle

Activity 9.7: Computing the length of the side BC in the figure below Working in pairs, revise how to find the length of the hypotenuse, BC, in the

figure below.

Figure 9.11: Right -angled triangle

Solution

Step 1: Using Pythagoras theorem a²= b²+ c²Step 2: Substitute values into the formula in step 1 BC²= 8²+ 6²Step 3: Simplify BC²= 64 + 36 = 100

Step 4: Find the square roots of both sides BC = √100 Therefore, BC = 10m Trigonometric ratios Trigonometric ratios are mathematical relationships between the sides of a right- angled triangle, defined as the ratios of the lengths of its sides relative to a specific acute angle.

Activity 9.8: Revision of the basic trigonometric ratios Working in pairs carry out the following activity to revise the basic trigonometric ratios.

Step 1: Sketch a right-angled triangle.

Step 2: Label all the three sides, as below.

Step 3: Determine the sine, cosine and the tangent of the acute angle, θ.

y x z θ

Figure 9.12: Right -angled triangle Hopefully you recalled that the three basic trigonometric ratios based on the triangle above are:

• sineθ = Opposite_________ Hypotenuse = y_ z, written as sinθ

• cosineθ = Adjacent_________ Hypotenuse = x_ z, written as cosθ

• tangentθ = Opposite_______ Adjacent = y_ x, written as tanθ For an easy reminder of the above trigonometric ratios, remember the following:

• Sine = Opposite_________ Hypotenuse = SOH

• Cosine = Adjacent_________ Hypotenuse = CAH

• Tangent = Opposite_______ Adjacent = TOA

• (SOH : CAH : TOA)

Example 9.10

In a right-angled triangle, the hypotenuse is 10 cm and one angle is 30⁰. Find the lengths of the other two sides.

Solution

sin 30 = Opp____ Hyp = ||AC|_ |BC| = ||AC|_ 10 |AC| = 10sin 30 = 10 × 0.5 = 5cm By Pythagoras’ theorem:

BC²= AB²+ AC²10²= AB²+ 5²AB²= 10²− 5²= 100 − 25 = 75 |AB| = √75 = 8.660 = 8.7cm Alternative Approach:

Cos30 = ||AB|_ |10| 10 × cos30 = |AB| |AB| = 10 × cos30 = 10 × 0.866 = 8.66 = 8.7cm The inverse of trigonometric ratios The inverse trigonometric ratios are cosecant, secant and cotangent.

y x z θ These are:

• cosecθ = 1___ sinθ = z_ y, therefore cosecθ = z_ y

• secθ = 1____ Cosθ = z_ x , therefore secθ = z_ x

• cotθ = 1____ tanθ = x_ y, therefore cotθ = x_ y

Activity 9.9: Derivation of the inverse of trigonometric ratios In small groups, work together to see how the above inverse trigonometric ratios were derived. Explain them to each other.

Solving problems with the trigonometric ratios

Example 9.11

If sin θ = 3/5 , find cos θ , tan θ and the value of θ in degrees.

5 3 x θ

Solution

By Pythagoras theorem 5²= 3²+ x²x²= 5²− 3²= 25 − 9 = 16 x = √16 = 4 Cosθ = 4/5 and tanθ = 3/4 Trigonometry in Real Life Areas in real life where trigonometry is applicable are:

1. Building Design, Architecture and construction: Roof angles and support structures; Bridge design; etc

2. Surveying and Flight Navigation: Calculating distances and heights; GPS and triangulation; etc.

3. Music: Analysis of sound waves and frequencies

4. Oceanography and Meteorology: Analysis of wave patterns and ocean currents.

Sample equipment used in trigonometry

Figure 9.13: Sample equipment used in trigonometry Angles of Elevation and Depression Angles of Elevation of an object B from an observer at A, who is below the level of B, is the angle which AB makes above the horizontal.

B A θ Angle of Elevation Horizontal

Figure 9.14: Angle of elevation

Example 9.12

A surveyor measures the angle of elevation to the top of a building to be 60⁰.

If the surveyor is standing 30 metres away from the base of the building what is the height of the building?

Solution

Step 1: Use the tangent ratio to relate the angle, the distance from the building and the height of the building: tan(60) = height/30

Step 2: Solve for the height:

tan(60) = height/30 Height = 30 × tan(60) = 51.96m Angles of Depression of an object Q from an observer at R, who is above the level of Q is the angle which RQ makes below the horizontal.

Angle of Depression

R Q θ Horizontal

Example 9.13

A person is standing on a cliff looking out at a ship in the distance. The angle of depression to the ship is 30⁰.

If the person is standing 20 metres above sea level, how far is the ship from the shore.

Solution

Use the tangent ratio to relate the angle, the height above sea level and the distance to the ship: tan(30⁰) = 20______ distance Solve for the distance to the ship:

Distance = 20______ tan(30) = 34.64m Examples 9.14 A boy is standing near a tree. He looks up at the tree and wonders, “How tall is the tree?”

What method could the boy use to find the height of the tree which does not involve climbing it?

Solution

We have a right-angled triangle.

Figure 9.15: Angles of elevation and depression In a right-angled triangle, ∆ABC , ta n of a ngl e θ is:

the ratio of the height of the tree to the distance between boy and foot of the tree.

∴ tanθ = Height of tree_____________________________ Horizontal distance between boy and tree = AB___ CB For example, if the distance |C B| = 30m and the angle formed θ = 45⁰, then tanθ = AB___ CB tan45° = B____ 30m Height, AB, = 30ta n × 45° = 30m

Example 9.15

A building designer wants to create a right angled, triangular roof with a 30° angle from the base to the sloping side. If the length of the base of the roof is 10 metres, what is the vertical height of the roof?

Solution

Step 1: Identify the given information:

The angle of the roof is 30° and the length of the base is 10 metres.

Step 2: Determine the trigonometric ratio to use:

Since we know the angle and the adjacent side (the base), we can use the tangent ratio to find the height.

Step 3: Set up the equation:

tan (30°) = Height/10

Step 4: Solve for the height:

height = 10 × tan (30°) = 5.773 metres Therefore, the height of the roof is 5.773m

Example 9.16

An airplane is flying at an altitude of 3000 metres. The angle of depression to an airfield on the ground is 20°.

If the plane were to start descending on the hypotenuse, what is the distance to the airfield to the nearest 10m?

Solution

Step 1: Identify the given information:

The altitude of the airplane is 3000 metres and the angle of depression is 20°.

Step 2: Determine the trigonometric ratio to use:

Since we know the angle and the opposite side (the altitude), we can use the sine ratio to find the distance of the hypotenuse.

Step 3: Set up the equation:

sin(20) = 3000______________ Distance to airfield

Step 4: Solve for the distance:

Distance to airfield = 3000_____ sin(20) = 8771 .4, Therefore, the distance the plane must fly along the hypotenuse is 8770m to the nearest 10m.

Example 9.17

A ladder leans against a brick wall. The foot of the ladder is 2m from the wall and makes a 30° angle with the ground.

How long is the ladder?

Solution

Step 1: Identify the given information:

The angle between the ground and the ladder is 60° and the distance from the ground to the wall is 1.5m.

Step 2: Determine the trigonometric ratio to use:

Since we know the angle and the adjacent side (the distance from the wall) and the ladder is the hypotenuse, we can use the cosine ratio to find the length of the ladder. .

Step 3: Set up the equation:

cos(60) = 1.5____________ Length of Ladder

Step 4: Solve for the length of ladder Length of ladder = 1.5______ cos(60) = 3m .

Example 9.18

A man is 20m from the base of a flagpole. His eyeline is 1.7m above the ground.

The angle of elevation from his eyeline to the top of the flagpole is 40°.

How tall is the flagpole?

Solution

Step 1: Identify the given information:

The angle between the man and the top of the flagpole is 40° and the distance from the man to the base of the flagpole is 20m.

Step 2: Determine the trigonometric ratio to use:

Since we know the angle and the adjacent side (the distance from the flagpole) and the flagpole is the opposite, we can use the tangent ratio to find the height of the flagpole from the man’s eyeline.

Step 3: Set up the equation:

tan(40) = Height from eyeline to top of flagpole/20

Step 4: Solve for the height from eyeline to top of flagpole:

Height from eyeline to top of flagpole = 20tan(40) = 16.78m.

Step 5: Calculate the full height of the flagpole:

We need to know the full height of the flagpole, so we must add on the height of the man’s eyeline.

Therefore, the full height of the flagpole = 16.78 + 1.7 = 18.48m

1. Find ⟶ PQ for the points P(− 2/3 ), Q(4/7).

2. A is (2/5), B is ( 6/12), C is (− 3/10 ) and D is( 8/19). Convert in component forms the vectors: ⟶ AB, ⟶ AC, ⟶ DA and ⟶ DB.

3. Given P (− 7/1 ), Q (− 3/6 ), and S ( 3___ − 3).

a. Find → PS and ⟶ PQ.

b. Calculate the lengths of → PS and ⟶ PQ.

4. The point A is (2, 3) and ⟶ AB = ( 4___ − 9), what are the coordinates of point B?

5. A person standing on the ground observes a bird on top of a tree at an angle of elevation of 60°. If the person is 30 metres away from the base of the tree, how high is the bird above the ground?

6. If s ecq = 17/8 find:

a. the trigonometric ratios of

(i) s inθ and

(ii) ta nθ

b. the value of q in degrees.

7. If ta nq = 15/8 , find the values of cos q and s inq.

8. If cotq = 2, find the values of s inq, cos q and ta nq .

9. If cos ecq = √_ 5, find the values sin q, cos q, ta nq and q in degrees.

10. A ladder is placed against a wall. The bottom of the ladder is 6 metres from the wall, and the top of the ladder touches the wall 8 metres above the ground. How long is the ladder?

11. A boat is 100 metres from the shore. The angle of elevation from the boat to the top of a lighthouse is 30°. How tall is the lighthouse?

12. A ladder, 10 metres long, leans against a wall, making an angle of 60° with the ground. How far is the base of the ladder from the wall?

13. A pilot descends toward a runway. The plane is flying at an altitude of 3 000 feet, and the angle of depression to the runway is 50⁰. What is the horizontal distance of the plane from the runway?

14. An engineer needs to calculate the length of a support cable for a tower.

The cable is attached to the top of a 50-metre tower and makes a 40⁰angle with the ground. How long is the cable?