Heights and distances: problems with two right triangles
When there are two angles, draw two right triangles that share a side. Write one tan equation for each, then solve them together.
Four common types
| Type | What is shared | The two equations |
|---|---|---|
| Observer walks towards a tower (two elevations) | The height of the tower | tan of each angle = height/(each distance) |
| Flagpole or tank on a building (two elevations from one point) | The distance from the building | One for the building, one for building + pole |
| From a building: elevation of a tower's top and depression of its foot | The horizontal distance | One above eye level, one below it |
| From a tall building: depressions of the top and foot of a shorter object | The horizontal distance | One for the whole building, one for the part above the object |
How to draw the figure (observer walking towards a tower)
- 1Draw the ground line and a vertical tower AB, with top A and foot B.
- 2Mark two points on the ground on the same side: D (nearer) and C (farther).
- 3Join AD and AC. Mark the larger angle at D and the smaller angle at C.
- 4Write CD (the distance walked), and let BD = x and AB = h.
- 5You now have two right triangles, ABD and ABC, sharing the side AB.
Example 1: walking towards a gopuram
A girl walks towards a temple gopuram. The angle of elevation of its top changes from 30° to 60° after she walks 20 m. Find the height of the gopuram. (√3 = 1.732)
- Formula: tan θ = opposite side/adjacent side
- Let AB = h, BD = x (the nearer point D) and CD = 20 m, so BC = x + 20.
- In triangle ABD: tan 60° = h/x, so h = √3 x ... (1)
- In triangle ABC: tan 30° = h/(x + 20), so h = (x + 20)/√3 ... (2)
- From (1) and (2): √3 x = (x + 20)/√3, so 3x = x + 20
- 2x = 20, so x = 10 m
- h = √3 × 10 = 10 × 1.732 = 17.32
The gopuram is 17.32 m high, and she is now 10 m from it.
Example 2: a flagpole on a building
A flagpole stands on top of a building 15 m high. From a point on the ground, the angles of elevation of the bottom and the top of the flagpole are 30° and 45°. Find the height of the flagpole. (√3 = 1.732)
- Formula: tan θ = opposite side/adjacent side
- Let BC = 15 m be the building (B foot, C top), CD = p the flagpole, and A the point on the ground with AB = d.
- Triangle ABC: tan 30° = 15/d, so d = 15√3 m
- Triangle ABD: tan 45° = (15 + p)/d, so 15 + p = d = 15√3
- p = 15√3 − 15 = 15(√3 − 1)
- p = 15 × 0.732 = 10.98
The flagpole is 10.98 m high.
Example 3: elevation and depression from a house
From the top of a house 10 m high, the angle of elevation of the top of a mobile tower is 60° and the angle of depression of its foot is 30°. Find the height of the tower.
- Formula: tan θ = opposite side/adjacent side
- Let the house be AB = 10 m (A top) and the tower CD (C top, D foot). Draw AE horizontal to meet the tower at E. Then ED = 10 m and AE = BD = d.
- Depression of the foot D is 30°: tan 30° = ED/AE = 10/d, so d = 10√3 m
- Elevation of the top C is 60°: tan 60° = CE/AE = CE/(10√3)
- CE = 10√3 × √3 = 30 m
- Height of the tower CD = CE + ED = 30 + 10 = 40 m
The mobile tower is 40 m high.
A surveyor who cannot reach the foot of a hill takes two angle readings from two spots a known distance apart. Two right triangles then give the height, for example when planning a ghat road or a mobile tower.
5 marks: “The angles of elevation of the top of a hill from two points on the same side, 100 m apart, are 30° and 45°. Find the height of the hill. (√3 = 1.732)” Nearer point: tan 45° = h/x, so x = h. Farther point: tan 30° = h/(h + 100), so h + 100 = √3 h. Then h = 100/(√3 − 1) = 50(√3 + 1) = **136.6 m**.
5 marks: “From the top of an 18 m high building, the angles of depression of the top and the bottom of a tree are 30° and 60°. Find the height of the tree.” Bottom: tan 60° = 18/d, so d = 6√3 m. Top: tan 30° = (18 − t)/d, so 18 − t = 6√3 × (1/√3) = 6. **Height of the tree = 12 m**.
Two angles, two triangles, one shared side.