Question

A man of height 6ft observes the top of a tower at angles of 45 and 30 of elevation and depression respectively. The height of tower is

Answer 1

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Answer 2

Answer:

16.392 ft

Step-by-step explanation:

Refer the attached figure

Height of man = AB = 6 ft

AB = ED = 6 ft

Let DC be x

Height of tower = ED+DC = 6+x

InΔ ABE

$tan \theta = \frac{Perpendicular}{Base}$

$tan 30^{\circ}= \frac{AB}{AE}$

$\frac{1}{\sqrt{3}}= \frac{6}{AE}$

$AE= \frac{6}{\frac{1}{\sqrt{3}}}$

$AE=10.392$

AE = BD = 10.392 ft

InΔ BCD

$tan \theta = \frac{Perpendicular}{Base}$

$tan 45^{\circ}= \frac{CD}{BD}$

$1= \frac{CD}{10.392}$

$10.392=CD$

Height of tower = ED+DC = 6+x = 6+10.392=16.392 ft

Hence Height of tower is 16.392 ft