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Find the height of the tower in each case.

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Let the height of the tower CD be x

Height of boy (AB) = CE

So, height of DE = CD – CE

= x – 1.6

In the triangle ΔDAE

∠DAE = 45°

From the trigonometric formula

$\tan \theta = \dfrac{{perpendicular}}{{base}}$

∵ Distance between the boy and tower

= 20 cm

So, BC = 20 cm

tan 45 $ = \dfrac{{x - 1.6}}{{20}}$

1 × 20 = x – 1.6 (∵ tan 45° = 1)

x = 21.6

tan 60° $ = \dfrac{{x - 1.6}}{{20}}$

$20\sqrt 3 + 1.6 = x$

x = 36.24

And the height of tower in case of elevation of 60° = 36.24.