Question

If a Tower 30m high casts a shadow $10\sqrt{3}m$ long on the ground, then what is the angle of elevation of the sun?


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Answer 1

Step-by-step explanation:

Answer:

• The angle of elevation of the sun is 60°.

• Step-by-step explanation:

Given : If a tower 30 m high casts a shadow $10\sqrt{3}$ m long on the ground.

To find : What is the angle of elevation of the sun?

Solution :

Let the position of sun is at 'A'.

• According to question a right angle triangle is form $△ABC$

With AB as height of the tower AB=30 m

and BC is the base where shadow cast $\rm BC=10\sqrt{3}$

• Let \theta be the angle of elevation.

Applying trigonometry property,

$\tt\tan\theta=\frac{AB}{BC}tan$

$\tt\tan\theta=\frac{30}{10\sqrt3}tanθ$

$\tt\tan\theta=\frac{3}{\sqrt3}$

$\tt\tan\theta=\sqrt3tanθ$

$\tt\theta=\tan^{-1}(\sqrt3)θ=tan$

$\tt\theta=\frac{\pi}{3}$

$\tt\theta=60^\circθ=60$

• Therefore, The angle of elevation of the sun is 60°.

Answer 2

Answer:

In triangle ABC :

Take height of tower as the perpendicular

Then the shadow as the base

So the solution is:

tan∅ = 30/10√3

tan∅ = 3/√3

tan∅ = √3 × √3 / √3  [ 3 = √3 × √3 ]

tan∅ = √3

tan∅ = tan 60°

∅ = 60°

Therefore the angle of elevation is 60°.

Hope it helps..

Step-by-step explanation:

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