Question

In a right triangle, one angle measures $x^o$, where $\sin x^o=\frac{4}{5}$. What is $\cos ({90}^o-x^o)$?

• A. $\frac{4}{5}$
• B. $\frac{3}{5}$
• C. $\frac{2}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{4}{5}$

Given, $\sin x^o=\frac{4}{5}$

$\Rightarrow \cos ({90}^o-x^o)=\sin \left(x^o\right)$

$\Rightarrow \cos ({90}^o-x^o)=\frac{4}{5}$

The other way through which this problem can be looked upon is that cosine of the third angle would be the ratio of the side adjacent to it and the hypotenuse.

And the side adjacent to the angle $({90}^o-x^o)$ is the one opposite to $x^o$.