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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 1 · Set 1 · Q37
Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna…
Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O' (as shown in figure).
Distance between the base of the tower and point ' O ' is $\displaystyle 6$ m . From point ' O ', the angle of elevation of the top of the section ' B ' is $\displaystyle 30^{\circ}$ and the angle of elevation of the top of section 'A' is $\displaystyle 60$°. Based on the above information, answer the following questions :
(i)
Find the length of the wire from the point 'O' to the top of section 'B'.
(ii)
Find the length of the wire from the point 'O' to the top of section 'A'.
(iii)
(a)
Find the distance AB.
OR
(b)
Find the area of $\displaystyle \triangle \mathrm{OPB}$.
Marking-scheme solution
(i) \(\displaystyle \cos 30^{\circ}=\frac{\sqrt{3}}{2}=\frac{6}{\mathrm{OB}}\)
\[\Rightarrow \mathrm{OB}=\frac{12}{\sqrt{3}} \text { or } 4 \sqrt{3} \mathrm{~m}
\]
(ii) \(\displaystyle \cos 60^{\circ}=\frac{1}{2}=\frac{6}{\mathrm{OA}}\)
\[\text { ⇒ OA = } 12 \text { m }
\]
(iii) (a) \(\displaystyle \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{BP}}{6}\)
\[\begin{aligned}
& \Rightarrow \mathrm{BP}=2 \sqrt{3} \mathrm{~m} \\
& \tan 60^{\circ}=\sqrt{3}=\frac{\mathrm{AP}}{6} \\
& \Rightarrow \mathrm{AP}=6 \sqrt{3} \mathrm{~m} \\
& \mathrm{AB}=\mathrm{AP}-\mathrm{BP}=6 \sqrt{3}-2 \sqrt{3}=4 \sqrt{3} \mathrm{~m}
\end{aligned}
\]
(iii) \[\begin{aligned}
& \text { (b) } \begin{aligned}
\tan 30^{\circ} & =\frac{1}{\sqrt{3}}=\frac{\mathrm{BP}}{6} \\
\Rightarrow \mathrm{BP}=2 & \sqrt{3} \mathrm{~m} \\
\operatorname{ar}(\triangle \mathrm{OPB}) & =\frac{1}{2} \times \mathrm{BP} \times \mathrm{OP} \\
& =\frac{1}{2} \times 2 \sqrt{3} \times 6=6 \sqrt{3} \mathrm{~m}^{2}
\end{aligned}
\end{aligned}
\]
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