Question

Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be 60°. Then she climbed a nearby observation deck. 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be 45°.

Based on the above given information, answer the following questions:

1. If CD is h metres, find the distance BD in terms of 'h'.
2. Find distance BC in terms of 'h'.
3. (a) Find the height CE of the lighthouse [Use `sqrt3` = 1.73]
   OR
4. (b) Find distance AE, if AC = 100m.

Answer 1

Given:

A lighthouse CE.
Amrita first at point A (ground). Angle of elevation to top C = 60°.
Then she goes to B, which is 40 m above A. Angle of elevation from B to top C = 45°.
BD = 40 m, CD = h, so total lighthouse height: CE = h + 40.

At point B, angle of elevation to C = 45°.

`tan 45° = (CD)/(DE)`

`1 = h/(DE)`

DE = h

At point A, angle of elevation to C = 60°.

`tan 60° = (CE)/(AE)`

But AE = DE (same horizontal distance) = h.

`sqrt3 = (h+40)/h`

Solve for h

`sqrt3 xx h = h + 40`

`(sqrt3 - 1)h = 40`

`h = 40/(sqrt3-1)`

Multiply numerator & denominator by `sqrt3 + 1`

`h = 40(sqrt3 + 1)/((sqrt3 - 1)(sqrt3 + 1))`

`h = (40(sqrt3+1))/(3-1)`

`h = 20 (sqrt3+1)`

h = 20(1.73 + 1) = 20(2.73) = 54.6m

(i) Distance BD in terms of h

BD is given = 40 m

(ii) Distance BC in terms of h

BC = BD + CD = 40 + h

(a) Find CE (height of lighthouse): CE = h + 40 = 54.6 + 40 = 94.6m

(b) If AC = 100 m, find AE:

AE² + CE² = AC²

`AE = sqrt(AC^2 - CE^2)`

`sqrt(100^2 - (94.6)^2)`

`AE = sqrt(10000-8944.16)`

`= sqrt1055.84`

= 32.5 m

BD = 40 m

BC = h + 40

(a) Height of lighthouse CE ≈ 94.6 m

(b) Distance AE ≈ 32.5 m