Let P sin- -cos- 0 - 2- be a point and let OAB be a triangle with vertices 0-0- -3-2 -0- and - 0-3-2- - The range of - if P lies inside the triangle OAB may be

The correct option is A ( 0,π/12) 0 lt; sinθ + cosθ #160; lt; √(32) .If sinθ + cosθ = √(32) . θ = π12 .From θ = 0 to π12 ...

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Byju's Answer

Standard XII

Physics

Introduction

Let P sinθ ...

Question

Let $P (sin θ, cos θ) (0 \leq θ \leq 2 π)$ be a point and let OAB be a triangle with vertices $(0, 0) [\sqrt{\frac{3}{2}}, 0]$ and $[0, \sqrt{\frac{3}{2}}]$ . The range of $θ$ if $P$ lies inside the triangle OAB may be

(0, \frac{π}{12})

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(\frac{5 π}{12}, \frac{π}{2})

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[π, \frac{5 π}{3}]

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[\frac{π}{2}, π]

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Solution

The correct option is A $(0, \frac{π}{12})$
$0 < sin θ + cos θ < \sqrt{\frac{3}{2}}$ .
If $sin θ + cos θ = \sqrt{\frac{3}{2}}$ .
$θ = \frac{π}{12}$ .
From $θ = 0$ to $\frac{π}{12}$ .
$sin θ + cos θ$ is increasing, So
$(0, \frac{π}{12})$ can be a part of range of $θ$ .
$A$ is correct.

Suggest Corrections

Similar questions

Q. Let

P (sin θ, cos θ)

, where

0 \leq θ \leq 2 π

be a point and let

O A B

be a triangle with vertices

(0, 0), (\frac{\sqrt{3}}{2}, 0)

and

(0, \frac{\sqrt{3}}{2})

. Find

θ

P

lies inside the

△ O A B

Q. Assertion :The value of

θ, 0 < θ < \frac{π}{2},

satisfying the equation

6 \sum m = 1 c o s e c [θ + \frac{(m - 1) π}{4}] c o s e c [θ + \frac{m π}{4}] = 4 \sqrt{2}

are

\frac{π}{12}

and

\frac{5 π}{12}

Reason: If

tan θ + cot θ = 4, 0 < θ < \frac{π}{2},

then

θ = \frac{π}{12}

\frac{5 π}{12}

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\frac{- 5 π}{12} \leq θ \leq - \frac{π}{3}

Max. Value of

cos (\frac{π}{6} + θ) - tan (θ + \frac{π}{6}) + tan (θ + \frac{2 π}{3})

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makes an angle

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with

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\frac{π}{4}

with

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and

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is :

Q. Let

P (sin θ, cos θ) (0 \leq θ \leq 2 π)

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Let P(sinθ,cosθ)(0≤θ≤2π) be a point and let OAB be a triangle with vertices (0,0)[√32,0] and [0,√32] . The range of θ if P lies inside the triangle OAB may be

The correct option is A (0,π12)0<sinθ+cosθ<√32.If sinθ+cosθ=√32.θ=π12.From θ=0 to π12.sinθ+cosθ is increasing, So (0,π12) can be a part of range of θ.A is correct.

Let $P (sin θ, cos θ) (0 \leq θ \leq 2 π)$ be a point and let OAB be a triangle with vertices $(0, 0) [\sqrt{\frac{3}{2}}, 0]$ and $[0, \sqrt{\frac{3}{2}}]$ . The range of $θ$ if $P$ lies inside the triangle OAB may be

The correct option is A $(0, \frac{π}{12})$
$0 < sin θ + cos θ < \sqrt{\frac{3}{2}}$ .
If $sin θ + cos θ = \sqrt{\frac{3}{2}}$ .
$θ = \frac{π}{12}$ .
From $θ = 0$ to $\frac{π}{12}$ .
$sin θ + cos θ$ is increasing, So
$(0, \frac{π}{12})$ can be a part of range of $θ$ .
$A$ is correct.