Question

From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be 45° and 60° respectively. If the height of the tower is 20 m. Find:

1. the height of the cliff.
2. the distance between the cliff and the tower.

Answer 1

Given: From the top of a cliff, the angles of depression to the top and bottom of a tower are 45° and 60° respectively; height of the tower = 20 m.

Step-wise calculation:

1. Let H = height of the cliff measured from the same horizontal level as the base of the tower and x = horizontal distance between the cliff and the tower.

2. Vertical drop from cliff-top to top of tower = H – 20.

Using angle of depression 45°:

`tan 45^circ = (H - 20)/x = 1`

⇒ x = H – 20 

3. Using angle of depression 60° to the bottom of the tower:

`tan 60^circ = H/x = sqrt(3)` 

⇒ `x = H/sqrt(3)`

4. Equate the two expressions for x:

`H - 20 = H/sqrt(3)` 

Multiply both sides by `sqrt(3)`:

`sqrt(3)H - 20sqrt(3) = H`

⇒ `H(sqrt(3) - 1) = 20sqrt(3)`

⇒ `H = (20sqrt(3))/(sqrt(3) - 1)`

5. Rationalize/simplify: 

`H = (20sqrt(3)(sqrt(3) + 1))/(3 - 1)`

= `10sqrt(3)(sqrt(3) + 1)`

= `10(3 + sqrt(3))`

= `30 + 10sqrt(3)`

Numerical: `sqrt(3) ≈ 1.732`

⇒ H = 30 + 17.32

⇒ H = 47.32 m (approx)

6. Distance x = H – 20

= `(30 + 10sqrt(3)) - 20` 

= `10 + 10sqrt(3)` 

= `10(1 + sqrt(3))` 

Numerical: x = 10(1 + 1.732)

= 10(2.732)

= 27.32 m

i. Height of the cliff

= `30 + 10sqrt(3)  m`

= 47.32 m

ii. Horizontal distance between the cliff and the tower

= `10(1 + sqrt(3)) m` 

= 27.32 m