Find the angular elevation of the sun when the shadow of a 10 m long pole is $10 \sqrt{3} m .$

$30^{\circ}$

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$60^{\circ}$

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$45^{\circ}$

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$15^{\circ}$

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Solution

The correct option is A
$30^{\circ}$

$Let the angle of elevation of sun = θ$
$\Rightarrow tan θ = \frac{P e r p e n d i c u l a r}{b a s e} = \frac{10}{10 \sqrt{3}}$
$\Rightarrow tan θ = \frac{1}{\sqrt{3}}$
$\Rightarrow θ = 30^{\circ}$

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Find the angle of elevation of the sun when the shadow of a 10m long pole is 10 $\sqrt{3}$ m.

Question 15
A pole 6m high casts a shadow $2 \sqrt{3} m$ long on the ground, then the Sun’s elevation is

(A) $60^{\circ}$
(B) $45^{\circ}$
(C) $30^{\circ}$
(D) $90^{\circ}$

Q. If the length of shadow a pole on a level ground is twice the length of the pole, the angle of elevation of the sun is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

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Find the angular elevation of the sun when the shadow of a 10 m long pole is 10√3 m.

The correct option is A 30∘ Let the angle of elevation of sun=θ ⇒tanθ=Perpendicularbase=1010√3 ⇒tanθ=1√3 ⇒θ=30∘

Find the angular elevation of the sun when the shadow of a 10 m long pole is $10 \sqrt{3} m .$

The correct option is A
$30^{\circ}$

$Let the angle of elevation of sun = θ$
$\Rightarrow tan θ = \frac{P e r p e n d i c u l a r}{b a s e} = \frac{10}{10 \sqrt{3}}$
$\Rightarrow tan θ = \frac{1}{\sqrt{3}}$
$\Rightarrow θ = 30^{\circ}$