(a)OR(b)

The angle of elevation of the top of a tower $\displaystyle 24$ m high from the foot of another tower in the same plane is $\displaystyle 60^{\circ}$. The angle of elevation of the top of second tower from the foot of the first tower is $\displaystyle 30^{\circ}$. Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.

Let AB and CD be the given towers. \[\begin{array}{l} \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{x}=\mathrm{h} \sqrt{3} \\ \tan 60^{\circ}=\sqrt{3}=\frac{24}{\mathrm{x}} \Rightarrow \mathrm{x}=\frac{24}{\sqrt{3}} \text { or } 8 \sqrt{3} \end{array} \] using (i) and (ii) \[\mathrm{x}=8 \sqrt{3} \text { and } \mathrm{h}=8 \] length of wire $\displaystyle =\sqrt{\mathrm{BE}^{2}+\mathrm{x}^{2}}=\sqrt{256+192}=\sqrt{448} \mathrm{~m}=8 \sqrt{7} \mathrm{~m}$ Let Point B represents observer. \[\therefore \angle \mathrm{QBP}=60^{\circ} ; \angle \mathrm{ABO}=45^{\circ} \] Using geometry $\displaystyle \angle \mathrm{PBO}=\frac{1}{2} \times 60^{\circ}=30^{\circ}$ Now, $\displaystyle \frac{\mathrm{r}}{\mathrm{OB}}=\sin 30^{\circ}=\frac{1}{2} \Rightarrow \mathrm{OB}=2 \mathrm{r}$ Also $\displaystyle \frac{\mathrm{OA}}{\mathrm{OB}}=\sin 45^{\circ}=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{OB}=\mathrm{OA} \sqrt{2}$ Using (i) and (ii) $\displaystyle \mathrm{OA}=\sqrt{2} \mathrm{r}$ or height of center of balloon $\displaystyle =\sqrt{2} \mathrm{r}$ units