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From the top of an upright pole 24 3 24\sqrt{3} 24 3 ​ feet high, the angle of elevation of the top of an upright tower was 60 ∘ 60^\circ 6 0 ∘ . If t...

SSC SSC CGL Tier 2 Medium Trigonometric Ratios MCQ

Question

From the top of an upright pole 24324\sqrt{3} feet high, the angle of elevation of the top of an upright tower was 60∘60^\circ. If the foot of the pole was $60$ feet away from the foot of the tower, what is the height (in feet) of the tower?

Options

A.

84384\sqrt{3}

B.

36336\sqrt{3}

C.

44344\sqrt{3}

D.

60360\sqrt{3}

trigonometryheight and distanceangle of elevationright triangle

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