Question

In $\Delta ABC,\angle C={90}^{\circ }$ then the value of ${\sin }^{2}A+{\sin }^{2}B$ is :

• A. $1$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $0$

Answer 1

The correct option is A $1$

Given $\angle C={90}^o$

From Triangle,

$\sin A=\frac{BC}{AB};$ $\sin B=\frac{AC}{AB}$

$\Rightarrow {\sin }^{2}A+{\sin }^{2}B$

$\frac{{BC}^2}{{AB}^2}+\frac{{AC}^2}{{AB}^2}=\frac{{BC}^2+{AC}^2}{{AB}^2}$

From $△ABC$, by using Pythagoras Theorem,

${AB}^2={BC}^2+{AC}^2$

${\sin }^{2}A+{\sin }^{2}B=\frac{{BC}^2+{AC}^2}{{AB}^2}=\frac{{AB}^2}{{AB}^2}=1$