Question

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

Answer 1

Let PAQ be the hill, AB be the tree and PCX be the horizontal. Let P be the point of observation. Produce BA to meet PX at C.

Let AB = h metres.

Then, $\angle XPA={15}^{\circ },PA=35m,$

$\angle CPB={60}^{\circ }and\angle PCA={90}^{\circ }.$

$\therefore \angle APB=({60}^{\circ }-{15}^{\circ })={45}^{\circ }.$

$In\Delta PAC,\angle PAC={180}^{\circ }-({15}^{\circ }+{90}^{\circ })={75}^{\circ }$

$\therefore \angle PAB=({180}^{\circ }-75)={105}^{\circ }$

And, $\angle PBA={180}^{\circ }-({45}^{\circ }+{105}^{\circ })={30}^{\circ }.$

Applying sine rule on $\Delta PAB,$ we get

$\frac{PA}{sin\angle PBA}=\frac{AB}{sin\angle APB}=\frac{35}{sin{30}^{\circ }}=\frac{H}{sin{45}^{\circ }}$

$\therefore 35\times 2=H\times \sqrt{2}\Rightarrow H=35\sqrt{2}m.$

Hence, the height of the trww is $35\sqrt{2}$m.