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in Trigonometric Functions by (15.7k points)

A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.

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Let BC represent the tree, A be the point 35m down the hill from the base of the tree and h be the height of the tree.

Clearly in ∆ABC

∠BAC = 60°- 15° =45°;

∠ACB = 30°; ∠ABC = 105°

\(\frac{AB}{sin 30 ^\circ}\) = \(\frac{BC}{sin 45 ^\circ}\)

\(\Rightarrow\) \(\frac{35}{\frac{1}{2}} = \frac{h}{\frac{1}{\sqrt 2}}\) \(\Rightarrow\) h = 49.5 m

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