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