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Question
Read the following passage:
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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 36 cm. From point O, the angle of elevation of the top of the Section B is 30° and the angle of elevation of the top of Section A is 45°.
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Based on the above information, answer the following questions:
- Find the length of the wire from the point O to the top of Section B.
- Find the distance AB.
OR
Find the area of ∠OPB. - Find the height of the Section A from the base of the tower.
Solution
i. Let the length of wire BO = x cm
∴ cos 30° = `(PO)/(BO)`
`\implies sqrt(3)/2 = 36/x`
`\implies` x = `(36 xx 2)/sqrt(3) xx sqrt(3)/sqrt(3)`
= `12 xx 2sqrt(3)`
= `24sqrt(3)` cm
ii. In ΔAPO, tan 45° = `(AP)/(PO)`
`\implies` 1 = `(AP)/36`
`\implies` AP = 36 cm ...(1)
Now, In ΔPBO,
tan 30° = `(BP)/(PO)`
`\implies 1/sqrt(3) = (BP)/36`
`\implies` BP = `36/sqrt(3) xx sqrt(3)/sqrt(3)`
= `36/3 sqrt(3)`
= `12sqrt(3)` cm
∴ AB = AP – BP
= `36 - 12sqrt(3)` cm
OR
In ΔOPB
tan 30° = `(BP)/(PO)`
`\implies 1/sqrt(3) = (BP)/36`
BP = `36/sqrt(3)`
= `36/sqrt(3) xx sqrt(3)/sqrt(3)`
= `12sqrt(3)` cm
Now, Area of ΔOPB = `1/2 xx "height" xx "base"`
= `1/2 xx BP xx OP`
= `1/2 xx 12sqrt(3) xx 36`
= `216sqrt(3)` cm2
iii. In ΔAPO, tan 45° = `(AP)/36`
1 = `(AP)/36`
A = 36 cm
Height of section A from the base of the tower = AP = 36 cm
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