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The shadow of a flagstaff is three times as long as the shadow of the flagstaff when the sun rays meet the ground at angle of $60^{o}$ . Find the angle between the sun rays and ground at the time of the longer shadow.

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θ = 30^{o}

θ = 60^{o}

θ = 45^{o}

θ = 15^{o}

answer is A.

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Detailed Solution

It is given that the shadow of a flagstaff is three times as long as the shadow of the flagstaff when the sun rays meet the ground at angle of

60^{o}

.
Let the length of the shadow be

x

when it makes an angle of

60^{0}

at point C at the ground and

3 x

at an angle of

θ

at point D.
Let height of flagstaff be

h

In right angled

ΔABC

\tan 60^{0} = \frac{h}{x}

⇒

h = x tan 60^{0}

..............(1)
In

ΔBDA,

tanθ = \frac{h}{3 x}

⇒

h = 3 xtanθ

...............(2)
From (1) & (2) we get,
⇒

xtan 60^{0} = 3 xtanθ

⇒

tanθ = \frac{\tan 6 0^{0}}{3}

⇒

tanθ = \frac{\sqrt{3}}{3}

⇒

tanθ = \frac{\sqrt{3} \times \sqrt{3}}{3 \times \sqrt{3}}

⇒

tanθ = \frac{1}{\sqrt{3}}

θ = 30^{0}

Hence, the correct option is 1.

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