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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 2 · Set 2 · Q35
(a) The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60^° and the angle of elevation of the…
(a)
The angle of elevation of the top of a tower $\displaystyle 30$ m high from the foot of another tower in the same plane is $\displaystyle 60^{\circ}$ and the angle of elevation of the top of the second tower from the foot of the first tower is $\displaystyle 30$°. Find the distance between the two towers and also the height of the other tower.
OR
(b)
From the top of a tower $\displaystyle 100$ m high, a man observes two cars on the opposite sides of the tower with angles of depression $\displaystyle 30$° and $\displaystyle 45^{\circ}$ respectively. Find the distance between the two cars. (Use $\displaystyle \sqrt{3}=1 \cdot 73$ )
Marking-scheme solution
(a) \(\displaystyle \mathrm{PQ}=\) height of \(\displaystyle 1^{\mathrm{St}}\) tower \(\displaystyle =30 \mathrm{~m} \mathrm{AB}=\) height of \(\displaystyle 2^{\mathrm{nd}}\) tower \(\displaystyle =\mathrm{h}\) (say)
\[\angle \mathrm{PAQ}=60^{\circ}, \angle \mathrm{APB}=30^{\circ}
\] Let AP = x
In \(\displaystyle \triangle \mathrm{APQ}, \tan 60^{\circ}=\frac{30}{\mathrm{x}} \Rightarrow \mathrm{x}=\frac{30}{\sqrt{3}}\)
\[=10 \sqrt{3}
\]
∴ Distance between two towers \(\displaystyle =10 \sqrt{3} \mathrm{~m}\)
In \(\displaystyle \triangle \mathrm{APB}, \tan 30^{\circ}=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{10 \sqrt{3}}\)\[\text { ⇒ h = } 10
\] ∴ Height of \(\displaystyle 2^{\text {nd }}\) tower \(\displaystyle =10 \mathrm{~m}\)
\(\displaystyle \mathrm{AB}=\) Height of tower \(\displaystyle =100 \mathrm{~m}\) P and Q are position of cars
\[\begin{aligned}
& \angle \mathrm{XBP}=\angle \mathrm{APB}=30^{\circ} \\
& \angle \mathrm{YBQ}=\angle \mathrm{AQB}=45^{\circ}
\end{aligned}
\] In \(\displaystyle \triangle \mathrm{ABQ}\),
\[\begin{aligned}
& \tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{AQ}} \Rightarrow 1=\frac{100}{\mathrm{x}} \\
& \Rightarrow \mathrm{x}=100
\end{aligned}
\] In \(\displaystyle \triangle \mathrm{ABP}\),
\[\begin{aligned}
& \tan 30^{\circ}=\frac{A B}{A P} \\
& \begin{aligned}
\frac{1}{\sqrt{3}} & =\frac{100}{y} \Rightarrow y=100 \sqrt{3} \\
& =100(1 \cdot 73)=173
\end{aligned}
\end{aligned}
\] Distance between cars \(\displaystyle =\mathrm{x}+\mathrm{y}\)
\[\text { = } 100 \text { + } 173 \text { = } 273
\] ∴ Distance between cars is $\displaystyle 273$ m.
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