Question

The angles of elevation of the top of a tower from two points at a distance of 4 m and
9 m from the base of the tower and in the same straight line with it are complementary
Prove that the height of the tower is 6 m.​

Answer 1

Height = 6  m

Step-by-step explanation:

The answer attached below explains the diagram and the procedure :

we are given angles of elevation at 2 different points from the base of the tower ,

2 right angles ACO and ABO can be used to proceed the questions .

2 angles a1 and a2  which are the angle of elevation are complementary ,

i,e

a1+a2= 90 degree

tan(a1+a2)  = Tan (90) = infinity

using the trigonometric formula of tan(a+b)  

we find that for tan(a1+a2) to be infinity

1 - tan (a1) * tan (a2)    = 0

rest is the equation modificati0n .

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Let the tower be AB.

Let the points be P and Q from where the angle of elevation forms a complementary angle.

So, let the angle of elevation of the top of a tower formed by point Q be :- $\alpha$.

And the angle of elevation of the top of a tower formed by point P be :- $90° - \alpha$.

The point P is 4 metres away from the base of the tower and the point Q is 9 metres away from the base of the tower.

We need to prove that the height of the tower is 6 metres.

Let us assume that the height is x.

• Now, let's solve !!!

From △ ABQ,

$\implies \tan( \alpha ) = \dfrac{x}{9}$

_______Let it be result number 1.

From △ ABP,

$\implies \tan(90 - \alpha ) = \dfrac{x}{4}$

Solving further,

$\implies \cot( \alpha ) = \dfrac{x}{4}$

$\implies \dfrac{1}{ \tan( \alpha ) } = \dfrac{x}{4}$

$\implies \tan( \alpha ) = \dfrac{4}{x}$

______Let this be result number 2.

Both the results have same LHS.

So, they can be equated.

Therefore,

$\implies \dfrac{x}{9} = \dfrac{4}{x}$

By cross multiplication,

$\implies {x}^{2} = 9 \times 4$

$\implies {x}^{2} = 36$

$\implies x = \sqrt{36}$

$\implies x = 6 \: m$

So, the height of the tower is 6 metres.

HENCE PROVED !!!

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