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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 1 · Set 1 · Q33
(a) The angle of elevation of the top of a tower 24 m high from the foot of another tower in the same plane is 60^°. The angle of elevation of the…
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.
A spherical balloon of radius r subtends an angle of $\displaystyle 60$° at the eye of an observer. If the angle of elevation of its centre is $\displaystyle 45$° from the same point, then prove that height of the centre of the balloon is $\displaystyle \sqrt{2}$ times its radius.
Marking-scheme solution
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
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