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(C) Let the length of the pole be $L$.The pole makes an angle of $60^{\circ}$ with the ground.At high noon,the sun is directly overhead,so the shadow

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$6$

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(C) Let the length of the pole be $L$.The pole makes an angle of $60^{\circ}$ with the ground.At high noon,the sun is directly overhead,so the shadow falls along the ground.The pole,its shadow,and the ground form a right-angled triangle where the pole is the hypotenuse and the shadow is the base.Using trigonometry,$\cos(60^{\circ}) = \frac{	ext{Base}}{	ext{Hypotenuse}} = \frac{	ext{Length of shadow}}{	ext{Length of pole}}$.$\cos(60^{\circ}) = \frac{3}{L}$.Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $\frac{1}{2} = \frac{3}{L}$.$L = 3 	imes 2 = 6 \, m$.Thus,the length of the pole is $6 \, m$.

$A$ pole stands slanting on the ground making an angle of $60^{\circ}$ with the ground. At high noon,the length of the shadow of the pole is $3 \, m$. Find the length of the pole in $m$.

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