If A and B are acute angles satisfying 3 cos^ | vedclass

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Question

(A) Given equations are:
$(1) 3 \cos^2 A + 2 \cos^2 B = 4$
$(2) \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A} \implies 3 \sin A \cos A = 2 \sin

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) Given equations are:
$(1) 3 \cos^2 A + 2 \cos^2 B = 4$
$(2) \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A} \implies 3 \sin A \cos A = 2 \sin B \cos B$
Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get:
$\frac{3}{2} \sin 2A = \sin 2B \implies 3 \sin 2A = 2 \sin 2B$
From $(1)$,$3 \cos^2 A + 2 \cos^2 B = 4$
$\implies 3(1 - \sin^2 A) + 2(1 - \sin^2 B) = 4$
$\implies 5 - 3 \sin^2 A - 2 \sin^2 B = 4$
$\implies 3 \sin^2 A + 2 \sin^2 B = 1$
Also,from $(2)$,$3 \sin A \cos A = 2 \sin B \cos B$. Squaring both sides:
$9 \sin^2 A \cos^2 A = 4 \sin^2 B \cos^2 B$
$\implies 9 \sin^2 A (1 - \sin^2 A) = 4 \sin^2 B (1 - \sin^2 B)$
Let $u = \sin^2 A$ and $v = \sin^2 B$. We have $3u + 2v = 1 \implies v = \frac{1 - 3u}{2}$.
Substituting into the squared equation:
$9u(1 - u) = 4(\frac{1 - 3u}{2})(1 - \frac{1 - 3u}{2})$
$\implies 9u - 9u^2 = 2(1 - 3u)(\frac{1 + 3u}{2})$
$\implies 9u - 9u^2 = 1 - 9u^2$
$\implies 9u = 1 \implies u = \frac{1}{9}$
Then $v = \frac{1 - 3(1/9)}{2} = \frac{1 - 1/3}{2} = \frac{2/3}{2} = \frac{1}{3}$.
So $\sin^2 A = \frac{1}{9} \implies \sin A = \frac{1}{3}$ and $\sin^2 B = \frac{1}{3} \implies \sin B = \frac{1}{\sqrt{3}}$.
Using $\cos^2 A = 1 - 1/9 = 8/9$ and $\cos^2 B = 1 - 1/3 = 2/3$.
For acute angles $A, B$,$\sin A = 1/3, \cos A = \sqrt{8}/3, \sin B = 1/\sqrt{3}, \cos B = \sqrt{2/3}$.
Using $\sin(A+2B) = \sin A \cos 2B + \cos A \sin 2B = \sin A (1 - 2\sin^2 B) + \cos A (2 \sin B \cos B)$
$= (1/3)(1 - 2/3) + (\sqrt{8}/3)(2 \cdot 1/\sqrt{3} \cdot \sqrt{2/3})$
$= (1/3)(1/3) + (2\sqrt{2}/3)(2\sqrt{2}/3) = 1/9 + 8/9 = 1$.
Thus,$A + 2B = \frac{\pi}{2}$.

If $A$ and $B$ are acute angles satisfying $3 \cos^2 A + 2 \cos^2 B = 4$ and $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$,then $A + 2B =$

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