A lamp post $5 \sqrt{3}$ m high casts a shadow 5 long on the ground. The Sun's elevation at this point is

30^{\circ}

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45^{\circ}

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60^{\circ}

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90^{\circ}

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Solution

The correct option is C $60^{\circ}$
Given,
Height of the post is $5 \sqrt{3}$ and base is $5$ m
$tan θ = \frac{A B}{B C}$
$∴ tan θ = \frac{5 \sqrt{3}}{5}$
$= \sqrt{3}$
∴ $∴ θ = 60^{0}$ ............ [since $tan 60^{0} = \sqrt{3}$ ]

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Similar questions

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 a pole 6 m high casts a shadow

2 \sqrt{3}

m long on the ground, then the Sun's elevation is ________.

Q. A pole 6 m high casts a shadow

2 \sqrt{3}

m long on the ground. The sun's altitude is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

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A lamp post 5√3 m high casts a shadow 5 long on the ground. The Sun's elevation at this point is

The correct option is C 60∘Given,Height of the post is 5√3 and base is 5mtanθ=ABBC∴tanθ=5√35 =√3∴∴θ=600 ............ [since tan600=√3]

A lamp post $5 \sqrt{3}$ m high casts a shadow 5 long on the ground. The Sun's elevation at this point is

The correct option is C $60^{\circ}$
Given,
Height of the post is $5 \sqrt{3}$ and base is $5$ m
$tan θ = \frac{A B}{B C}$
$∴ tan θ = \frac{5 \sqrt{3}}{5}$
$= \sqrt{3}$
∴ $∴ θ = 60^{0}$ ............ [since $tan 60^{0} = \sqrt{3}$ ]