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

30°

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60 °

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45°

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15°

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Solution

The correct option is A
30°

Let, 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 10m long pole is 10√3m.

The correct option is A 30° Let, 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 10m long pole is 10 $\sqrt{3}$ m.

The correct option is A
30°

Let, 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 $θ$ = $\frac{1}{\sqrt{3}}$
$\Rightarrow$ $θ$ = $30^{\circ}$