If the angle of elevation of an object from a point 200 meters above the lake is found to be 300 and the angle of depression of its image in the lake is 450, then the height of the object above the lake is
A
200(√3−1)(√3+1) meters
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B
200(√3−1)√3 meters
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C
200(√3+1)√3meters
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D
200(√3+1)(√3−1) meters
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Solution
The correct option is D200(√3+1)(√3−1) meters
Let the height of the object above the lake is ′h′m.
Hence, the reflection of object in the lake will be at the same distance ′h′m from the surface of lake.
Let AD=BC=xm
Also, AB+DC+200m
DE=(h−200)
In, △ADE,
⇒tan30°=EDDA=(h−200)x
⇒1√3=(h−200)x
⇒x=√3(h−200)⟶(1)
In, △ADF,
⇒tan45°=FDDA=(h+60)x
⇒1=(h+60)x
⇒x=h+200⟶(2)
From (1)ξ(2), we get
⇒h+200=√3h−200√3
⇒√3h−h=200√3+200
⇒(√3−1)h=200(√3+1)
⇒h=200(√3+1)(√3−1)
Hence, height of object from the lake is 200(√3+1)(√3−1)m.