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The angles of depression of two points A and B on a horizontal

Question

The angles of depression of two points A and B on a horizontal plane such that AB = 200 from the top P of a tower PQ of height 100 are $4 5^{0} - Θ$ and $4 5^{0} + Θ$ . If the line AB passes through Q the foot of the tower, then angle $Θ$ is equal to

4 5^{0}

3 0^{0}

22. 5^{0}

1 5^{0}

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The angles of depression of two points A and B on a horizontal plane such that AB = 200 from the top P of a tower PQ of height 100 are

4 5^{0} - Θ

and

4 5^{0} + Θ

. If the line AB passes through Q the foot of the tower, then angle

Θ

is equal to

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Text solutionVerified

AB = 200 = AQ -BQ
200 = 100cot (

4 5^{0} - θ

) - 100cot (

4 5^{0} + θ

)
200 = 100

[\frac{1 + t an θ}{1 - t an θ} - \frac{1 - t an θ}{1 + t an θ}]

2 =

\frac{4 t an θ}{1 - t a n ^{2} θ} = t an 2 θ

θ = 22. 5^{0}

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Question Text	The angles of depression of two points A and B on a horizontal plane such that AB = 200 from the top P of a tower PQ of height 100 are $4 5^{0} - Θ$ and $4 5^{0} + Θ$ . If the line AB passes through Q the foot of the tower, then angle $Θ$ is equal to
Answer Type	Text solution:1
Upvotes	150