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In a $\Delta PQR$ ,

if $3\sin P+4\cos Q=6\:\: and\: \: 4\sin Q+3\cos P=1$ , then the angle R is equal to

Option 1)
$\frac{5\pi }{6}$

Option 2)
$\frac{\pi }{6}$

Option 3)
$\frac{\pi }{4}$

Option 4)
$\frac{3\pi }{4}$

Answers (1)

As we learnt in

Trigonometric Equations -

The equations involving trigonometric function of unknown angles are known as trigonometric equations.

- wherein

e.g. $\cos ^{2}\Theta - 4\cos \Theta = 1$

$3 \sin P+4\cos Q= 6$

$4\sin Q+3\cos P= 1$

Squaring both sides, we get

$9sin^{2}P+16cos^{2}Q+24\ sinP\ cos\ Q+16 sin^{2}Q+9 cos^{2}P+24\ sin\Q\ cos\ P=37$

$\left \Rightarrow (sin^{2}P+cos^{2}P)+16(sin^{2}Q+cos^{2}Q)\\ +24(sinP\ cos\ Q+sin\ Q\ cos\ P)=37$

$\Rightarrow 25+24\ sin(P + Q)=37$

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

$\Rightarrow \sin (\pi-(R))=\frac{1}{2}$

$R=\frac{\pi}{6}\ \, \, \, or\ \, \, \, \frac{5\pi}{6}$

If $R= \frac{5\pi }{6},then\ \: \: P< \frac{\pi }{6}\Rightarrow 3\sin P< \frac{1}{2} \Rightarrow sin P<\frac{3}{2}$

Thus $4 cos Q=6-\frac{3}{2}=\frac{9}{2}\Rightarrow cosQ > 1, not\ possible$

$\Rightarrow R=\frac{\pi }{6}$

Option 1)

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

Incorrect

Option 2)

$\frac{\pi }{6}$

Correct

Option 3)

$\frac{\pi }{4}$

Incorrect

Option 4)

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

Incorrect

Posted by

Vakul

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