Math, asked by ashwinesh09, 10 months ago

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

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Answers

Answered by tanyasingh99004
9

Step-by-step explanation:

here is your answer brother.

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Answered by Anonymous
32

Given : BC = 4 cm, BD = 9cm.

\angle{ACB} = (90° - Ø) and \angle{ADB} = Ø

Prove : Height of twoer i.e. AB = 6m

Proof :

Remember..

tanØ = \dfrac{P}{B} and cotØ = \dfrac{B}{P}

In \triangleABC

=> tan (90° - Ø) = \dfrac{AB}{BC}

=> cotØ = \dfrac{AB}{BC}

=> \dfrac{1}{tan\theta} = \dfrac{BC}{AB}

=> tanØ = \dfrac{BC}{AB} ____ (eq 1)

In \triangleABD

=> tanØ = \dfrac{AB}{BD} _____ (eq 2)

On comapring (eq 1) and (eq 2) we get

=> \dfrac{BC}{AB} = \dfrac{AB}{BD}

Cross multiply them

=> BC × BD = AB × AB

=> 4 × 9 = AB²

=> 36 = AB²

=> AB = 6 m

Height of tower 6 m.

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