There is a small island in between a river 100 meters wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are 30° and 45° respectively, find the height of the tree.

Let XY be tree of h meters.

From figure,

Considering right angled triangle △XQY,

$\Rightarrow \text{tan 45°} = \dfrac{XY}{YQ} \\[1em] \Rightarrow 1 = \dfrac{h}{YQ} \\[1em] \Rightarrow YQ = h \text{ ….(Eq 1)}$

Considering right angled triangle △XPY,

$\Rightarrow \text{tan 30°} = \dfrac{XY}{PY} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - YQ} \\[1em] \Rightarrow 100 - YQ = h\sqrt{3}$

Putting value of YQ from Eq 1 in above equation,

$\Rightarrow 100 - h = 1.732h \\[1em] \Rightarrow 1.732h + h = 100 \\[1em] \Rightarrow 2.732h = 100 \\[1em] \Rightarrow h = \dfrac{100}{2.732} = 36.6 \text{ m}.$

Hence, the height of tree is 36.6 m.