Question

Q1) An observer 1.2m tall is 28.2 m away from a tower. The angle of
elevation of the top of the tower from his eye is 60°. What is the height of
the tower?

Answer 1

Question:-

An observer 1.2m tall is 28.2 m away from a tower. The angle of elevation of the top of the tower from his eye is 60°. What is the height of the tower?

Solution:-

Given,

Height of observer = $1.2m$

Distance = $28.2m$

Elevation angle = $60°$

Let the height of the tower be $h \: metres$

In right-angled ∆CDE

$\implies{Tan \: E = \frac{CD}{DE}}$

$\implies{Tan \: 60° = \frac{h - 1.2}{28.2 m}}$

$\implies{\sqrt{3}}$ = $\frac{h - 1.2}{28.2}$

${\implies{h = 28.2\sqrt{3} + 1.2m}}$

$\bold{\red{\implies{h = 50.04m}}}$

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Answer 2

Answer:

33.76 m

Step-by-step explanation:

Given :

• Height of observer = 1.2 m
• Distance from Tower = 28.2 m
• Angle of elevation = 60°

$cos \: x = adj \div hyp$

• Adj = 28.2 m
• x = 60°

Cos 60° = 28.2 / hyp

1/2 = 28.2 / hyp

hyp = 28.2 × 2

hyp = 56.4 m

Height of tower above observer height = 56.4 m

Height of tower from ground = 56.4 + 1.2 m

Total Height of the tower = 57.6 m