Question

The angles $\alpha ,\beta ,\gamma$ of a triangle satisfy the equations

$2\sin \alpha +3\cos \beta =3\sqrt{2}$ and $3\sin \beta +2\cos \alpha =1$ .

Then angle $\gamma$ equals to?

• A. ${135}^o$
• B. ${120}^o$
• C. ${60}^o$
• D. ${30}^o$

Answer 1

The correct option is D ${30}^o$

$2\sin \alpha +3\cos \beta =3\sqrt{2}...\left(1\right)$

$3\sin \beta +2\cos \alpha =1...\left(2\right)$

Squaring and adding equations (1) and (2), we have

$4({\sin }^2\alpha +{\cos }^2\alpha )+9({\cos }^2\beta +{\sin }^2\beta )+12(\sin \alpha \cos \beta +\cos \alpha \sin \beta )=(3\sqrt{2}{)}^2+1^2=19$

$\therefore 4+9+12(\sin \alpha \cos \beta +\cos \alpha \sin \beta )=19$

$\therefore sin\alpha \cos \beta +\cos \alpha \sin \beta =\frac{1}{2}$

$\therefore \sin (\alpha +\beta )=\frac{1}{2}$

$\Rightarrow \sin (180-\gamma )=\sin \gamma =\frac{1}{2}$

$\therefore \gamma ={30}^o$