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AIEEE 2012 Solved Paper

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Question : 61 of 89

Marks: +1, -0

In a

\Delta P Q R

3 \sin P + 4 \cos Q = 6

4 \sin Q + 3 \cos P = 1

, then the angle R is equal to:

$\frac{5\pi}{6}$
$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{3\pi}{4}$

Solution:

Given

3 \sin P + 4 \cos Q = 6

4 \sin Q + 3 \cos P = 1

Squaring and adding

(i) \& (i i)

we get

9 \sin^2 P + 16 \cos^2 Q + 24 \sin P \cos Q

+16 \sin^2 Q + 9 \cos^2 P + 24 \sin Q \cos P

=36+1=37

\Rightarrow 9 ( \sin^2 p + \cos^2 P ) + 16 ( \sin^2 Q + \cos^2 q )

+24( \sin P \cos Q + \cos P \sin Q )=37

\Rightarrow 9+16+24 \sin (P+Q)=37

[ As

\sin^2 \theta + \cos^2 \theta = 1

and

\sin A \cos B + \cos A \sin B = \sin (A+B) ]

\Rightarrow \sin (P+Q) = \frac{1}{2}

\Rightarrow P+Q = \frac{\pi}{6} \text{ or } \frac{5\pi}{6}

\Rightarrow R = \frac{5\pi}{6} \text{ or } \frac{\pi}{6}

\text{(as } P+Q+R=\pi \text{)}

If

R = \frac{5\pi}{6}

then

0 < P, Q < \frac{\pi}{6}

\Rightarrow \cos Q < 1 \text{ and } \sin P < \frac{1}{2}

\Rightarrow 3 \sin P + 4 \cos Q < \frac{11}{2}

which is not true.

So

R = \frac{\pi}{6}

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