Question

Help Mehh

A tower subtends an angle a at a point A in the plane of its base and the angle of
depression of the foot of the tower at a point b metres just above Ais B. Prove that the
height of the tower is b tan a cot ß.​

Answer 1

Answer:

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Answer 2

HEY MATE YOUR ANSWER IS

HERE.....

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ATQ

IN TRIANGLE ABC

$\tan( \beta ) = \frac{ab}{ac}$

$ab = \: b$

$ab \: = \: \frac{b}{ \tan( \beta ) }$

$\frac{1}{ \tan( \beta ) } = \cot( \beta )$

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$ac = b \times \cot( \beta ) \: \: eq1$

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NOW IN TRIANGLE ADC

$\tan( \alpha ) = \frac{cd}{ac}$

$cd = ac \times \tan( \alpha )$

NOW PUT VALUE OF AC FROM Eq 1.............

$b \times \cot( \beta ) \times \tan( \alpha )$

hence .....

$b \tan( \alpha ) \cot( \beta )$

HENCE PROVED.....

THANKS FOR THE QUESTION

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