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Mathematics · 2023 · 4 marks
CBSE 2023 · Region 6 · 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 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. Distance between the base of the tower and point O is $\displaystyle 36$ cm. From point O, the angle of elevation of the top of the Section B is $\displaystyle 30$° and the angle of elevation of the top of Section A is $\displaystyle 45$°.
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 distance AB.OR Find the area of $\displaystyle \triangle \mathrm{OPB}$.(iii)Find the height of the Section A from the base of the tower.
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. Distance between the base of the tower and point O is $\displaystyle 36$ cm. From point O, the angle of elevation of the top of the Section B is $\displaystyle 30$° and the angle of elevation of the top of Section A is $\displaystyle 45$°.
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 distance AB.
OR Find the area of $\displaystyle \triangle \mathrm{OPB}$.
(iii)
Find the height of the Section A from the base of the tower.
Marking-scheme solution
(i)
In \(\displaystyle \triangle \mathrm{OBP}, \cos 30^{\circ}=\frac{O P}{O B}\)
\[\begin{aligned}
& \frac{\sqrt{3}}{2}=\frac{36}{O B} \Rightarrow O B=\frac{72}{\sqrt{3}} \\
& =24 \sqrt{3} \mathrm{~cm}
\end{aligned}
\]
(ii)
In \(\displaystyle \Delta \mathrm{OBP}, \tan 30^{\circ}=\frac{P B}{36} \Rightarrow \mathrm{~PB}=\frac{36}{\sqrt{3}}\)
\[\mathrm{PB}=12 \sqrt{3}
\]
In \(\displaystyle \triangle \mathrm{OAP}, \tan 45^{\circ}=\frac{A P}{36} \Rightarrow \mathrm{AP}=36 \mathrm{~cm}\)
\[A B=A P-P B=36-12 \sqrt{3}=12(3-\sqrt{3}) \mathrm{cm}
\]
(ii)
Area of \(\displaystyle \Delta \mathrm{OPB}=\frac{1}{2} \times \mathrm{OP} \times \mathrm{PB}\)
\[=\frac{1}{2} \times 36 \times 12 \sqrt{3}=216 \sqrt{3} \mathrm{~cm}^{2}
\]
(iii)
\(\displaystyle \mathrm{AP}=36 \mathrm{~cm}\)
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.