Question

In $△ABC$, $\angle B={90}^{\circ }$, $\sin C=\frac{3}{5}$, then $\cos A=$ ______

• A. $\frac{3}{5}$
• B. $\frac{4}{5}$
• C. $\frac{5}{4}$
• D. $\frac{5}{3}$

Answer 1

Answer: (A)

$\frac{3}{5}$

In a right angled triangle, $A+B+C={180}^{\circ }$

$B={90}^{\circ }$

So, $A+C={90}^{\circ }$

$A={90}^{\circ }-C$

$\cos A=\cos ({90}^{\circ }-A)$

$\Rightarrow \cos A=\sin C=\frac{3}{5}$

$\Rightarrow \cos A=\frac{3}{5}$