Question

The angle of elevation of the top of a tree is found to be 33° at one point and 59° at a top a point 31 ft. nearer the tree. How high is the tree if both observation points and the base of the tree are in the same horizontal plane?

Answer 1

Question 1.

The angle of elevation of the top of a tree is found to be $33{}^{\circ}$ at one point and $59{}^{\circ}$ at a point $31$ ft. nearer the tree. How high is the tree if both observation points and the base of the tree are in the same horizontal plane?

Solution. Let $A$ be the top of the tree, $B$ its bottom, $C_{1}$ and $C_{2}$ the points from which the top is seen under the angles $33{}^{\circ}$ and $59{}^{\circ}$, respectively. It is given that $\angle AC_{1}B=33{}^{\circ}$, $\angle AC_{2}B=59{}^{\circ}$ and $C_{1}C_{2}=31$. Therefore, $\angle AC_{2}C_{1}=180{}^{\circ}-59{}^{\circ}=121{}^{\circ}$ and $\angle C_{1}AC_{2}=180{}^{\circ}-121{}^{\circ}-33{}^{\circ}=26{}^{\circ}$. Use the law of sines for $\triangle AC_{1}C_{2}$:

$\frac{C_{1}C_{2}}{\sin\angle C_{1}AC_{2}}=\frac{AC_{2}}{\sin\angle AC_{1}C_{2}},$

and hence

$AC_{2}=C_{1}C_{2}\frac{\sin\angle AC_{1}C_{2}}{\sin\angle C_{1}AC_{2}}=31\frac{\sin 33{}^{\circ}}{\sin 26{}^{\circ}}.$

Then from $\triangle ABC_{2}$ we have

$AB=AC_{2}\sin\angle AC_{2}B=31\frac{\sin 33{}^{\circ}}{\sin 26{}^{\circ}}\cdot\sin 59{}^{\circ}\approx 33.$

Answer: approximately $33$ ft.