Test Index

AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Mathematics

© examsnet.com

Question : 24 of 160

Marks: +1, -0

In

\triangle A B C

3 \sin A+4 \cos B=6

4 \sin B+3 \cos A=1

, then the angle

C

is

$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$

Solution: 👈: Video Solution

3 \sin A+ 4 \cos B=6 \dots(i)

4 \sin B+ 3 \cos A=1 \dots(ii)

On adding

Eqs. (i)

and

(ii)

after taking square,

(9 \sin^2 A+24 \sin A \cdot \cos B+16 \cos ^2 B) +

(16 \sin^2 B+24 \sin B \cdot \cos A+9 \cos^2 A) = 37

\Rightarrow 9 ( \sin^2 A + \cos^2 A) +16 (\cos ^2 B+ \sin^2 B) + 24 \sin (A + B) = 37

\Rightarrow 9+16+24 \sin (A+B)=37

\Rightarrow 24 \sin (A+B)=12

\Rightarrow \sin (A+B)=\frac{1}{2}

\Rightarrow A+ B=\frac{\pi}{6}, \frac{5\pi}{6}

\because A+ B> C

\therefore A+ B= \frac{5\pi}{6}

Hence,

C=\pi-\frac{5\pi}{6}=\frac{\pi}{6}

\dots[ \because \ A+ B+ C=\pi]

© examsnet.com

Go to Question:

Prev Question

Next Question