Question

From the top of a building $AB,60m$ high, the angle of the depression of the top and bottom of 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 the lamp post.
(iii) The difference between the height of the building and the lamp post.

Answer 1

Consider the given data.

The height of building $AB=60$ m.

And angle of the depression of the top and bottom of vertical lamp post CD are observed to be ${30}^{\circ }$ and ${60}^{\circ }$, respectively shown in the figure.

In the $△ADB$,

$\tan {60}^{\circ }=\frac{60}{x}$

$\sqrt{3}=\frac{60}{x}$

$x=\frac{60}{\sqrt{3}}$ m

In the $△ACP$,

$\tan {30}^{\circ }=\frac{60-h}{x}$

$\frac{1}{\sqrt{3}}=\frac{60-h}{\frac{60}{\sqrt{3}}}$

$\frac{60}{3}=60-h$

$h=60-20$

$h=40$ m

Therefore,

The distance of between AB and CD is $\frac{60}{\sqrt{3}}$ m.

The height of lamp post is $40$ m.

The difference between the height of the building and the lamp post $=60-40=20$ m.