Question

25- The angle of depression of the top and the bottom of a 5m tall building from the top of a muiti-storeyed building are
30'and 45°respectively. Find the height of the multi-storeyed building and the distance between the two building.​

Answer 1

ANSWER:-

Given:

The angle of depression of the top and the bottom of a 5m tall building from the top of a multi-storeyed building are 30° & 45° respectively.

To find:

Find the height of the multi-storeyed building and the distance between the two building.

Solution:

Let PC denoted the multi-storeyed building & AB demoted the 5m tall building.

Given,

∠PBD= 30°

∠PAC= 45°

Let PD= h & AC= BD= x

Now,

In ∆BPD,

$= > tan30 \degree = \frac{PD}{BD} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{h}{x} \\ \\ = > x = \sqrt{3} h...............(1)$

&

In ∆DAC,

$= > tan45 \degree = \frac{PC}{AC} \\ \\ = > 1 = \frac{h + 5}{x} \\ \\ = > x = h + 5............(2)$

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

$= > \sqrt{3} h = h + 5 \\ \\ = > \sqrt{3} h - h = 5 \\ \\ = > h( \sqrt{3} - 1) = 5\\ \\ = > h = \frac{5}{ \sqrt{3} - 1 } \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1} \\ \\ = >h = \frac{5( \sqrt{3} + 1) }{( { \sqrt{3} )}^{2} - ( {1)}^{2} } \\ \\ = > h = \frac{5(\sqrt{3} + 1) }{3 - 1} \\ \\ = > h = \frac{5( \sqrt{3} + 1) }{2}$

Now,

Total height of multi-storeyed building:

$= > h + 5 \\ \\ = > \frac{5( \sqrt{3} + 1) }{2} + 5 \\ \\ = > 5( \sqrt{3} + 1) + 10 \\ \\ = > 5( \sqrt{3 } + 1 + 2) \\ \\ = > 5( \sqrt{3} + 3)$

Similarly,

Distance between two buildings(x);

=) 5(√3+3)m

Hope it helps ☺️