Question
Find A, if $0^\circ \leq A \leq 90^\circ$ and $4sin^2A - 3 = 0$
✓
Answer
$4 \sin ^2 A -3=0$
$\Rightarrow \sin ^2 A=\frac{3}{4}$
$\Rightarrow \sin A=\frac{\sqrt{3}}{2}$
We know $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
Hence, $A=60^{\circ}$
$\Rightarrow \sin ^2 A=\frac{3}{4}$
$\Rightarrow \sin A=\frac{\sqrt{3}}{2}$
We know $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
Hence, $A=60^{\circ}$
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