Some Applications of Trigonometry · Learner pages

Lesson 1 · Model one angle of elevation

Choose the right triangle and a suitable trigonometric ratio.

Name or initials: ____________   Date: ____________

An angle of elevation is measured upward from a horizontal line, not from the vertical.

1. Draw what is measured. A vertical pole stands on level ground. From a ground-level point 12 m away, its top has elevation 45°.
2. Label the right triangle. The horizontal leg is 12 m, the opposite vertical leg is h, and the angle at the observer is 45°.
3. Use tangent. tan 45° = h/12. Since tan 45° = 1, h = 12 m.
4. Check observation height. This model assumes observation at ground level. If an instrument is above ground, its height must be added to the calculated rise.

Worked example: At horizontal distance 10√3 m, a pole top has elevation 30°. h = 10√3 × tan 30° = 10√3/√3 = 10 m for a ground-level observer.

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.

   1. 5 m · 2. 10 m · 3. 2.5 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.

   1. 16 m · 2. 4 m · 3. 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.

   1. 5 m · 2. 10 m · 3. 20 m

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

Lesson 2 · Account for eye height and depression

Translate an observation into the correct vertical difference.

Name or initials: ____________   Date: ____________

Two horizontal lines are parallel, so a line of sight creates equal alternate interior angles.

1. Separate total height and rise. An instrument stands 1.5 m above ground, 20 m horizontally from a building. Its elevation to the top is 45°.
2. Calculate the rise. The right triangle gives rise = 20tan45° = 20 m.
3. Restore the observation height. Total building height = 20 + 1.5 = 21.5 m.
4. Relate depression to elevation. From the top looking down along the same line of sight, the depression below its horizontal equals the elevation above the instrument horizontal.

Worked example: A lookout 18 m above a level path sees a marker at depression 45°. The corresponding elevation at the marker is 45°, so horizontal separation = 18/tan45° = 18 m.

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.

   1. 11.5 m · 2. 10 m · 3. 8.5 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.

   1. 12 m · 2. 10.4 m · 3. 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.

   1. 6.6 m · 2. 9.4 m · 3. 8 m

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

Lesson 3 · Combine observations from two positions

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

Name or initials: ____________   Date: ____________

Moving closer to a fixed vertical tower on level ground increases its angle of elevation.

1. Name the nearer distance. Two ground-level observation points and the tower base lie on one straight level path. Both points are on the same side of the tower, with elevations 60° nearby and 30° from a point 20 m farther away. Let the nearer distance be x.
2. Write each height expression. h = x tan60° = x√3. Also h = (x + 20)tan30° = (x + 20)/√3.
3. Equate the same height. x√3 = (x + 20)/√3. Multiply by √3 to get 3x = x + 20; hence x = 10.
4. Find and verify the height. h = 10√3 m. The farther distance is 30 m, and 30/√3 = 10√3, so both observations agree.

Worked example: If the same 60°/30° observations are 12 m apart on one side, 3x = x + 12 gives x = 6 m and height 6√3 m.

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°?

   1. 3 m · 2. 9 m · 3. 6 m

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°?

   1. 12 m · 2. 8 m · 3. 4 m

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.

   1. 16 m · 2. 4 m · 3. 8 m

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

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.

   30 m / 7.5 m / 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.

   9 m / 18 m / 36 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.

   20 m / 18.3 m / 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.

   4.8 m / 7.2 m / 6 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.

   12 m / 24 m / 6 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 m / 8√3 m / 16√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 / 40 m / 10 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.

   48 m / 12 m / 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 / 18 m / 16.2 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.

   25 m / 23.5 m / 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 / 20 m / 40 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.

   24√3 m / 12 m / 12√3 m