Question

The angle of elevation of a tower is observed to be 45° at the end of horizontal base of 120 m measured from its feet. Find height of tower.

Answer 1

✬ Height = 120 m ✬

Step-by-step explanation:

Given:

• Angle of elevation of the tower is 45°.
• Measure of the base is 120 m.

To Find:

• What is the height of the tower ?

Solution: Let AB be the tower and let C be the position of observer. Therefore

In right angled ∆ABC,

• CB = 120 m
• ∠ABC = 90°
• ∠ACB = 45°

Let AB (Height of tower) be h metres.

From right ∆ABC we have

$\implies{\rm }$ tanθ = Perpendicular/Base

$\implies{\rm }$ tan45° = AB/CB

$\implies{\rm }$ 1 = h/120

$\implies{\rm }$ 120 $\times$ 1 = h

$\implies{\rm }$ 120 m = h

Hence, the height of the tower is AB = 120 m.

Answer 2

GIVEN:

• The angle of elevation of a tower is observed to be 45°
• The distance from B to C = 120 m

TO FIND:

• What is the height of the tower ?

SOLUTION:

Let the height of the tower be 'h' m

According to question:-

We know that,

$\rm{\dashrightarrow tan \theta = \dfrac{Perpendicular}{Base}}$

$\rm{\dashrightarrow tan \theta = \dfrac{AB}{BC}}$

We have given that, the angle of elevation of a tower is observed to be 45°

$\rm{\dashrightarrow tan 45\degree = \dfrac{AB}{BC}}$

We know, that the value of tan 45° is 1

$\rm{\dashrightarrow 1 = \dfrac{h}{120}}$

$\rm{\dashrightarrow 1 \times 120 = h}$

$\bf{\dashrightarrow 120 \: m = h}$

❝ Hence, the height of the tower is 120 m ❞

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✰ IDENTITIES ✰

➜ sin²θ + cos²θ = 1

➜ sec²θ = 1 + tan²θ

➜ cosec²θ = 1 + cot²θ