Question

5
.
36. A flag-staff of height h stands on the top of a school building. If the angles of des
of the top and bottom of the flag - staff have measure c. and B respectively from a
h tan 8
on the ground. Prove that the height of the building is tan
ng tana-tanß​

Answer 1

Answer:

It is proved that  height of building is  $\dfrac{h tan\beta }{tan\alpha - tan\beta }$ .

Step-by-step explanation:

Given as :

The height of flag staff = CD = h

The angle of elevation at top of staff = α

The angle of elevation at bottom of staff = β

To prove that , Height of building = $\dfrac{h tan\beta }{ tan\alpha - tan\beta }$

According to question

From figure

In Δ ABD

Tan β = $\dfrac{AB}{BD}$

Or, BD = AB cotβ                   ........1

Again

In Δ CDB

Tanα = $\dfrac{BC}{BD}$

Or, BD = BC cotα

Or, BD = (h + AB) cotα           ........2

From eq 1 and eq 2

AB cotβ =  (h + AB) cotα  

Or, AB cotβ = h cotα + AB cotα  

Or, AB cotβ - AB cotα = h cotα

Or, AB = $\dfrac{h cot\alpha }{ cot\beta - cot\alpha }$

Or, AB = $\dfrac{\frac{h}{tan\alpha }}{\frac{1}{tan\beta }- \frac{1}{tan\alpha }}$

Or, AB = $\dfrac{h tan\beta }{tan\alpha - tan\beta }$

So, The height of building = AB = $\dfrac{h tan\beta }{tan\alpha - tan\beta }$

Hence, It is proved that  height of building is  $\dfrac{h tan\beta }{tan\alpha - tan\beta }$ . Answer