Question

the angle of depression of the top and bottom of a 50 m high building from the top of a tower are 45 and 60 find the height of tower​

Answer 1

Answer:

This is what you want...

Step-by-step explanation:

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Answer 2

ANSWER:-

Given:

The angle of depression of the top & bottom of a 50m high building from the top of a tower are 45° & 60° respectively.

To find:

The height of the tower.

Solution:

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

•Draw CL perpendicular to AB.

So,

⚫LB= CD= 50m

⚫AL= (h-50)m

⚫LC = BD

Let angle ACL= 45° &

angle ADB= 60°

In ∆ABD,

We know, [tan60° = √3]

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

In ∆ALC,

We know [tan45°= 1]

$= > tan45 \degree = \frac{AL}{LC} \\ \\ = > 1 = \frac{h - 50}{BD} \\ \\ = > BD = h - 50.................(2)$

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

$= > h - 50 = \frac{h}{ \sqrt{3} } \\ \\ = > h - \frac{h}{ \sqrt{3} } = 50 \\ \\ = > \sqrt{3} h - h = 50 \sqrt{3} \\ \\ = > h( \sqrt{3} - 1) = 5 0\sqrt{3} \\ \\ = > h = \frac{50 \sqrt{3} }{ \sqrt{3} - 1} \\ [Rationalising] \\ \\ = > \frac{50 \sqrt{3} \times \sqrt{3} + 1 }{ \sqrt{3} - 1 \times \sqrt{3} + 1 } \\ \\ = > \frac{150 + 50 \sqrt{3} }{ { (\sqrt{3}) }^{2} - ( {1)}^{2} } \: \: \: \: \: \: [{a}^{2} - {b}^{2} = (a + b)(a - b)] \\ \\ = > \frac{150 + 50 \sqrt{3} }{3 - 1} \\ \\ = > \frac{150 + 50 \sqrt{3} }{2} \\ \\ = > 75 + 25 \sqrt{3} \\ \\ = > 75 + 25\times 1.732 \\ \\ = > (75 + 43.3)m \\ \\ = >h = 118.3m$

Thus,

The height of the tower is, h= 118.3m.

Hope it helps ☺️