Question

please solve this question...​

Answer 1

Given,

$\longrightarrow\sf{2\sin A+\sqrt3\sin B=\dfrac{5}{2}\quad\quad\dots(1)}$

$\longrightarrow\sf{\sqrt3\sin A+2\sin B=\dfrac{3\sqrt3}{2}\quad\quad\dots(2)}$

Adding (1) and (2),

$\longrightarrow\sf{2\sin A+\sqrt3\sin B+\sqrt3\sin A+2\sin B=\dfrac{5+3\sqrt3}{2}}$

$\longrightarrow\sf{\left(2+\sqrt3\right)\sin A+\left(2+\sqrt3\right)\sin B=\dfrac{5+3\sqrt3}{2}}$

$\longrightarrow\sf{\sin A+\sin B=\dfrac{5+3\sqrt3}{2(2+\sqrt3)}}$

$\longrightarrow\sf{\sin A+\sin B=\dfrac{5+3\sqrt3}{4+2\sqrt3}\quad\quad\dots(3)}$

Subtracting (2) from (1),

$\longrightarrow\sf{2\sin A+\sqrt3\sin B-\sqrt3\sin A-2\sin B=\dfrac{5-3\sqrt3}{2}}$

$\longrightarrow\sf{\left(2-\sqrt3\right)\sin A-\left(2-\sqrt3\right)\sin B=\dfrac{5-3\sqrt3}{2}}$

$\longrightarrow\sf{\sin A-\sin B=\dfrac{5-3\sqrt3}{2(2-\sqrt3)}}$

$\longrightarrow\sf{\sin A-\sin B=\dfrac{5-3\sqrt3}{4-2\sqrt3}\quad\quad\dots(4)}$

On adding (3) and (4),

$\longrightarrow\sf{2\sin A=\dfrac{5+3\sqrt3}{4+2\sqrt3}+\dfrac{5-3\sqrt3}{4-2\sqrt3}}$

$\longrightarrow\sf{2\sin A=\dfrac{(5+3\sqrt3)(4-2\sqrt3)+(5-3\sqrt3)(4+2\sqrt3)}{(4+2\sqrt3)(4-2\sqrt3)}}$

$\longrightarrow\sf{2\sin A=\dfrac{20-10\sqrt3+12\sqrt3-18+20+10\sqrt3-12\sqrt3-18}{16-12}}$

$\longrightarrow\sf{2\sin A=\dfrac{20-18+20-18}{16-12}}$

$\longrightarrow\sf{2\sin A=\dfrac{4}{4}}$

$\longrightarrow\sf{2\sin A=1}$

$\longrightarrow\sf{\sin A=\dfrac{1}{2}}$

Since A is an acute angle,

$\Longrightarrow\sf{A=30^o}$

Subtracting (4) from (3),

$\longrightarrow\sf{2\sin B=\dfrac{5+3\sqrt3}{4+2\sqrt3}-\dfrac{5-3\sqrt3}{4-2\sqrt3}}$

$\longrightarrow\sf{2\sin B=\dfrac{(5+3\sqrt3)(4-2\sqrt3)-(5-3\sqrt3)(4+2\sqrt3)}{(4+2\sqrt3)(4-2\sqrt3)}}$

$\longrightarrow\sf{2\sin B=\dfrac{20-10\sqrt3+12\sqrt3-18-20-10\sqrt3+12\sqrt3+18}{16-12}}$

$\longrightarrow\sf{2\sin B=\dfrac{-10\sqrt3+12\sqrt3-10\sqrt3+12\sqrt3}{16-12}}$

$\longrightarrow\sf{2\sin B=\dfrac{4\sqrt3}{4}}$

$\longrightarrow\sf{2\sin B=\sqrt3}$

$\longrightarrow\sf{\sin B=\dfrac{\sqrt3}{2}}$

Since B is also an acute angle,

$\Longrightarrow\sf{B=60^o}$

Here A, B and C are angles of a triangle. Hence,

$\longrightarrow\sf{A+B+C=180^o}$

$\longrightarrow\sf{C=180^o-A-B}$

$\longrightarrow\sf{C=180^o-30^o-60^o}$

$\longrightarrow\underline{\underline{\sf{C=90^o}}}$

Hence 90° is the answer.