From the top of a building, the angle of elevation of the top of a cell tower is $45^{o}$ and the angle of depression to foot is $60^{o}$ . If height of cell tower is $46 m e t e r s$ then find the distance between building and tower, and height of the building?

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Solution

$L e t t h e d i s t a n c e = B Q = x m$

$I n t h e Δ P Q B,$

$t a n 45^{o} = \frac{P Q}{B Q}$

$1 = \frac{h}{x}$

$x = h ⟹ (1)$

$I n t h e Δ P Q A,$

$t a n 60^{o} = \frac{P Q}{A Q}$

$1.732 = \frac{h}{45 + x}$

$1.732 (45 + h) = h$

$77.94 + 1.732 h = 20 ⟹ (2)$

$1.732 h = 20 - 77.94$

$h = \frac{- 57.94}{1.732}$

$S i n c e d i s t a n c e c a n t b e n e g a t i v e,$

$h = 33.45 m$

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