Question

From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp-post CD are observed to be ${30}^{\circ }$ and ${60}^{\circ }$ respectively. Find

(i) the horizontal distance between AB and CD,

(ii) the height of the lamp-post,

(iii) the difference between the heights of the building and the lamp-post.

Answer 1

Let AB be the building of height 60m and CD be the lamp post of height h, an angle of depression of the top and bottom of vertical lamp post is ${30}^o$ and ${60}^o$ respectively.

Let AE = h, AC = x and AC = ED

It is also given AB = 60 m, Then BE = 60-h and

$\angle ACB={60}^o,\angle BDE={30}^o$

(i) So we use trigonometric ratios,

In right $△ABC,\Rightarrow \tan {60}^o=\frac{AB}{AC}\Rightarrow \sqrt{3}=\frac{60}{x}\Rightarrow x=\frac{60}{\sqrt{3}}\Rightarrow x=34.64m$

Hence distance between AB and CD is 34.64 m

(ii) So again in right $△BDE,\Rightarrow \tan {30}^o=\frac{BE}{DE}\Rightarrow \frac{1}{\sqrt{3}}=\frac{60-h}{x}\Rightarrow x=(60-h)\sqrt{3}\Rightarrow \frac{60}{\sqrt{3}}=(60-h)\sqrt{3}\Rightarrow 60=180-3h\Rightarrow 3h=180-60\Rightarrow 3h=120\Rightarrow h=40m$

Hence height of the lamp post is 40 m

(iii) Since $BE=60-h\Rightarrow BE=60-40\Rightarrow BE=20$

Hence the difference between the height of building and lamp post is 20 m