Question
In $\triangle ABC , \sqrt{2} AC = BC , \sin A =1, \sin ^2 A +\sin ^2 B +\sin ^2 C =2$ then $\angle A =? \angle B =$ ? $\angle C =$ ?
✓
Answer
$\sin A=1$
But, $\sin 90^{\circ}=1$
$\therefore \sin A=\sin 90^{\circ}$
$\therefore A=90^{\circ}$
$\sqrt{2} AC = BC \ldots$....[Given]
$\therefore \frac{ AC }{ BC }=\frac{1}{\sqrt{2}}$
$\therefore \sin B=\frac{A C}{B C}$
(ii) [By definition]
$\therefore \sin B=\frac{1}{\sqrt{2}}$
But, $\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
$\therefore \sin B=\sin 45^{\circ}$
$\therefore B=45^{\circ}$
$\sin ^2 A +\sin ^2 B +\sin ^2 C =2$
$ \therefore(1)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\sin ^2 C =2$
$\therefore 1+\frac{1}{2}+\sin ^2 C =2$
$\therefore \sin ^2 C =2-\frac{3}{2} $
$\therefore \sin ^2 C =\frac{1}{2}$
$\therefore \sin C=\frac{1}{\sqrt{2}}$
But, $\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
$\therefore \sin C=\sin 45^{\circ}$
$\therefore C =45^{\circ}$
$\therefore \angle A=90^{\circ}, \angle B=45^{\circ}, \angle C=45^{\circ}$
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