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A vertical pole subtends an angle ^{-1}(1/2) at point P on the

Question

A vertical pole subtends an angle $tan^{- 1} (1/2)$ at point $P$ on the ground. The angle subtended by the upper half of the pole at the point $P$ is-

tan^{- 1} (1/4)

tan^{- 1} (2/9)

tan^{- 1} (1/8)

tan^{- 1} (2/3)

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A vertical pole subtends an angle

tan^{- 1} (1/2)

at point

P

on the ground. The angle subtended by the upper half of the pole at the point

P

is-

15 mins ago

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Text solutionVerified

∠ A BP = 9 0^{\circ}

and

∠ A PB = t a n^{- 1} (\frac{1}{2})

(given)

Let height of the pole be

h

Δ A BP

\frac{A B}{BP} = tan (∠ A PB)

\Rightarrow \frac{h}{BP} = tan (tan^{- 1} \frac{1}{2})

\Rightarrow \frac{h}{BP} = \frac{1}{2}

\Rightarrow BP = 2 h

---(i)

Δ D BP

\frac{D B}{BP} = tan (∠ D PB)

\Rightarrow \frac{\frac{h}{2}}{2 h} = tan (∠ D PB)

(from (i))

\Rightarrow \frac{h}{4 h} = tan (∠ D PB)

\Rightarrow tan (∠ D PB) = \frac{1}{4}

\Rightarrow ∠ D PB = tan^{- 1} (\frac{1}{4})

Angle subtended by upper half of the pole at point

P

= ∠ A P D = ∠ A PB - ∠ D PB = tan^{- 1} (\frac{1}{2}) - tan^{- 1} (\frac{1}{4})

= tan^{- 1} ⎝ ⎛ \frac{\frac{1}{2} - \frac{1}{4}}{1 - \frac{1}{2} \times \frac{1}{4}} ⎠ ⎞

= tan^{- 1} (\frac{2}{9})

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Question Text	A vertical pole subtends an angle $tan^{- 1} (1/2)$ at point $P$ on the ground. The angle subtended by the upper half of the pole at the point $P$ is-
Answer Type	Text solution:1
Upvotes	150