Question

In a $△PQR$, if 3 sinP + 4 cosQ = 6 and 4 sinQ + 3 cosP = 1, then the angle R is equal to:

• A. $\frac{3\pi }{4}$
• B. $\frac{5\pi }{6}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $\frac{3\pi }{4}$

We have,

$3\sin P+4\cos Q=6......\left(1\right)$

$4\sin Q+3\cos P=1.......\left(2\right)$

On squaring and adding both side and we get,

${(3\sin P+4\cos Q)}^2+{(4\sin Q+3\cos P)}^2=36+1$

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

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

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

$\Rightarrow 9\times 1+16\times 1+24\sin (P+Q)=37$

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

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

$\Rightarrow 24\sin (P+Q)=12$

$\Rightarrow \sin (P+Q)=\frac{12}{24}$

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

$\Rightarrow \sin (P+Q)=\sin {30}^0$

$\Rightarrow \sin (P+Q)=\sin {150}^0$

Now, $(P+Q)={30}^o,{150}^o$

Now $P+Q={30}^o$

So,

$P+Q+R={180}^o$

${30}^o+R={180}^o$

$R={150}^o=\frac{3\pi }{4}$

Hence, this is the answer