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प्रश्न
From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects.
उत्तर
Given: the height of tower is h m.
∠ABD = α and ∠ACD = β
Let CD = y and BC = x
In ∆ABD,
tan α = `"AD"/"BD"`
⇒ tan α = `"h"/("BC" + "CD")`
⇒ tan α = `"h"/(x + y)`
⇒ x + y = `"h"/tan α`
⇒ y = `"h"/tan α - x` ...[Equation 1]
In ∆ACD,
tan β = `"AD"/"CD"`
⇒ tan β = `"h"/y`
⇒ y = `"h"/tan β` ...[Equation 2]
Comparing equation 1 and equation 2,
`"h"/tan α - x = "h"/tan β`
⇒ x = `"h"/tan α - "h"/tan β`
⇒ x = `"h"(1/tan α - 1/tan β)`
⇒ x = h(cot α – cot β)
Hence, we have got the required distance between the two points, i.e. h(cot α – cot β)
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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.
