Question

If a tower 30 metre high , casts a shadow 10√3 metre long on the ground , then what is the angle of elevation of the sun ?

30°

60°

45°

90°​

Answer 1

✬ Angle of Elevation = 60° ✬

Step-by-step explanation:

Given:

• Height of tower is 30 m.
• Shadow of tower from its foot is 10√3 m.

To Find:

• What is the angle of elevation of the sun ?

Solution: Let angle of incidence be θ and AB be a tower and BC be the distance between foot of tower and shadow.

Here in right angled triangle ABC we have

• AB = 30 m (Perpendicular)
• BC = 10√3 m (Base)

We know that

★ tanθ = Perpendicular/Base ★

$\implies{\rm }$ tanθ = AB/BC

$\implies{\rm }$ tanθ = 30/10√3

$\implies{\rm }$ tanθ = 3/√3

• {now rationalising denominator}

$\implies{\rm }$ tanθ = 3/√3 × √3/√3

$\implies{\rm }$ tanθ = 3√3/3 = √3

• As we know that tan60° = √3

Hence, value of θ or angle of elevation of sun will be 60°.

Answer 2

Answer:

Question.

If a tower 30 metre high , casts a shadow 10√3 metre long on the ground , then what is the angle of elevation of the sun ?

30°

60°

45°

90

GIVEN

• Height of tower = 30 m
• Shadow cast = 10√3

To find

Angle of elevation of the sun

Solution

Let angle of incidence be θ and AB be a tower and BC be the distance between foot of tower and shadow.

So, here is a right angle triangle ABC we have

Perpendicular = AB = 30 m

Base = BC = 10√3

➡️tan∅ = Perpendicular /Base

➡️tan∅ = 30/10√3

➡️tan∅ = 3/√3

Rationalization of denominator

➡️tanθ = 3/√3 × √3/√3

➡️tanθ = 3√3/3 = √3

We know that √3 = 60⁰

$\huge \fbox {The answer is 60⁰}$