Some Applications of Trigonometry · Answer guide

Lesson 1 · Model one angle of elevation

Choose the right triangle and a suitable trigonometric ratio.

1. From ground level on flat ground, the top of a vertical pole has angle of elevation 45°. The foot is 5 m away. Find the height.

   Answer: 5 m. tan 45° = height/distance = 1, so height = 5 m.

2. From ground level on flat ground, the top of a vertical pole has angle of elevation 45°. The foot is 8 m away. Find the height.

   Answer: 8 m. tan 45° = height/distance = 1, so height = 8 m.

3. From ground level on flat ground, the top of a vertical pole has angle of elevation 45°. The foot is 10 m away. Find the height.

   Answer: 10 m. tan 45° = height/distance = 1, so height = 10 m.

Discuss: Sketch a 60° observation from 4 m away and express the pole height exactly before using a decimal approximation.

Teaching note: The sloping line of sight is not the horizontal distance. Decide which length the question actually gives. Open responses need discussion; they are not automatically marked.

Lesson 2 · Account for eye height and depression

Translate an observation into the correct vertical difference.

1. An instrument is 1.5 m above flat ground. At horizontal distance 10 m, elevation to a building top is 45°. Find the building height.

   Answer: 11.5 m. Rise above the instrument is 10tan45° = 10 m. Add its height: 10 + 1.5 = 11.5 m.

2. An instrument is 1.6 m above flat ground. At horizontal distance 12 m, elevation to a building top is 45°. Find the building height.

   Answer: 13.6 m. Rise above the instrument is 12tan45° = 12 m. Add its height: 12 + 1.6 = 13.6 m.

3. An instrument is 1.4 m above flat ground. At horizontal distance 8 m, elevation to a building top is 45°. Find the building height.

   Answer: 9.4 m. Rise above the instrument is 8tan45° = 8 m. Add its height: 8 + 1.4 = 9.4 m.

Discuss: Draw two parallel horizontal lines and a sloping sight line. Mark the equal elevation and depression angles.

Teaching note: An angle of depression is measured from the horizontal at the observer. It is not the angle between the sight line and a vertical wall. Open responses need discussion; they are not automatically marked.

Lesson 3 · Combine observations from two positions

Express the same height twice and solve for the unknown distance.

1. A tower of height 3√3 m is viewed from ground level on a straight level path. What is the nearer horizontal distance when the elevation is 60°?

   Answer: 3 m. tan60° = (3√3)/d = √3, so d = 3 m. A 30° observation would be 9 m away.

2. A tower of height 4√3 m is viewed from ground level on a straight level path. What is the nearer horizontal distance when the elevation is 60°?

   Answer: 4 m. tan60° = (4√3)/d = √3, so d = 4 m. A 30° observation would be 12 m away.

3. Two ground-level points lie on the same straight level path through a vertical tower base, on the same side of the tower, and are 8 m apart. Elevations are 60° at the nearer point and 30° at the farther point. Find the nearer horizontal distance.

   Answer: 4 m. Let nearer distance be x. The shared height gives x√3 = (x + 8)/√3, so 3x = x + 8. Thus x = 4 m and height = 4√3 m.

Discuss: Keep the same angles and double the separation. Explain why both the nearer distance and tower height double.

Teaching note: When both observation points are on the same side, the farther distance is x plus the separation. A diagram with points on opposite sides requires a different relation. Open responses need discussion; they are not automatically marked.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. From ground level on flat ground, the top of a vertical pole has angle of elevation 45°. The foot is 15 m away. Find the height.

   15 m. tan 45° = height/distance = 1, so height = 15 m.

2. From ground level on flat ground, the top of a vertical pole has angle of elevation 45°. The foot is 18 m away. Find the height.

   18 m. tan 45° = height/distance = 1, so height = 18 m.

3. An instrument is 1.7 m above flat ground. At horizontal distance 20 m, elevation to a building top is 45°. Find the building height.

   21.7 m. Rise above the instrument is 20tan45° = 20 m. Add its height: 20 + 1.7 = 21.7 m.

4. An instrument is 1.2 m above flat ground. At horizontal distance 6 m, elevation to a building top is 45°. Find the building height.

   7.2 m. Rise above the instrument is 6tan45° = 6 m. Add its height: 6 + 1.2 = 7.2 m.

5. Two ground-level points lie on the same straight level path through a vertical tower base, on the same side of the tower, and are 12 m apart. Elevations are 60° at the nearer point and 30° at the farther point. Find the nearer horizontal distance.

   6 m. Let nearer distance be x. The shared height gives x√3 = (x + 12)/√3, so 3x = x + 12. Thus x = 6 m and height = 6√3 m.

6. Two ground-level points lie on the same straight level path through a vertical tower base, on the same side of the tower, and are 16 m apart. Elevations are 60° at the nearer point and 30° at the farther point. Find the tower height.

   8√3 m. Let nearer distance be x. The shared height gives x√3 = (x + 16)/√3, so 3x = x + 16. Thus x = 8 m and height = 8√3 m.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. From ground level on flat ground, the top of a vertical pole has angle of elevation 45°. The foot is 20 m away. Find the height.

   20 m. tan 45° = height/distance = 1, so height = 20 m.

2. From ground level on flat ground, the top of a vertical pole has angle of elevation 45°. The foot is 24 m away. Find the height.

   24 m. tan 45° = height/distance = 1, so height = 24 m.

3. An instrument is 1.8 m above flat ground. At horizontal distance 18 m, elevation to a building top is 45°. Find the building height.

   19.8 m. Rise above the instrument is 18tan45° = 18 m. Add its height: 18 + 1.8 = 19.8 m.

4. An instrument is 1.5 m above flat ground. At horizontal distance 25 m, elevation to a building top is 45°. Find the building height.

   26.5 m. Rise above the instrument is 25tan45° = 25 m. Add its height: 25 + 1.5 = 26.5 m.

5. Two ground-level points lie on the same straight level path through a vertical tower base, on the same side of the tower, and are 20 m apart. Elevations are 60° at the nearer point and 30° at the farther point. Find the nearer horizontal distance.

   10 m. Let nearer distance be x. The shared height gives x√3 = (x + 20)/√3, so 3x = x + 20. Thus x = 10 m and height = 10√3 m.

6. Two ground-level points lie on the same straight level path through a vertical tower base, on the same side of the tower, and are 24 m apart. Elevations are 60° at the nearer point and 30° at the farther point. Find the tower height.

   12√3 m. Let nearer distance be x. The shared height gives x√3 = (x + 24)/√3, so 3x = x + 24. Thus x = 12 m and height = 12√3 m.