Question

Please solve this question

Answer 1

ANSWER:-

Given:

The angles of depression of the top & bottom of a building 50m high as observed from the top of a tower are 30° & 60° respectively.

To find:

The height of the tower & also the horizontal distance between the building & the tower.

Solution:

Let AB be the tower of height h m & CD be the building of height 50m.

⚫Draw CL perpendicular AB.

Now,

⚫LB= CD= 50m.

⚫AL= (h-50)m

⚫LC = BD

Let angle ALC =30° & angle ADB= 60°

In ∆ABD,

$tan60 \degree = \frac{AB}{BD} \\ \\ = > \sqrt{3} = \frac{h}{BD} \\ \\ = > \sqrt{3} BD = h \\ \\ = > BD = \frac{h}{ \sqrt{3} } .................(1)$

In ∆ALC,

$tan30 \degree = \frac{AL}{LC} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{h - 50}{BD} \\ \\ = > BD = \sqrt{3} (h - 50)..............(2)$

From equation (1) & (2), we get;

$\sqrt{3} (h - 50) = \frac{h}{ \sqrt{3} } \\ \\ = > \sqrt{3} h - 50 \sqrt{3} = \frac{h}{ \sqrt{3} } \\ \\ = > \sqrt{3} h - \frac{h}{ \sqrt{3} } = 50 \sqrt{3} \\ \\ = > 3h - h = 50 \times 3 \\ \\ = > 2h = 150 \\ \\ = > h = \frac{150}{2} \\ \\ = > h = 75m$

Putting the value of h in equation (2), we get;

=) BD= √3(75 - 50)

=) BD= 1.732( 25)

=) BD= 43.3m

So,

⚫Height of the tower, h= 75m.

⚫Horizontal distance between tower & building is 43.3m.

Hope it helps ☺️