Question

If ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=9/4,$ then

• A. $\angle A+\angle B=\angle C$
• B. $\angle A+\angle B+\angle C=0$
• C. $\angle A=\angle B=\angle C={60}^{\circ }$
• D. $\angle A+\angle B={90}^{\circ }-\angle C$

Answer 1

The correct option is C $\angle A=\angle B=\angle C={60}^{\circ }$

Given equation

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=\frac{9}{4}$

Put $\angle A=\angle B=\angle C={60}^{\circ }$

${\sin }^{2}60+{\sin }^{2}60+{\sin }^{2}60$

$⟹3\times {\left(\frac{\sqrt{3}}{2}\right)}^2$

$⟹3\times \frac{3}{4}=\frac{9}{4}$

$\therefore \angle A=\angle B=\angle C={60}^{\circ }$