Question

If $cos{25}^0+sin{25}^0=k$, then $cos{20}^0$ is equal to

• A. $\frac{k}{\sqrt{2}}$
• B. $-\frac{k}{\sqrt{2}}$
• C. $\pm \frac{k}{\sqrt{2}}$
• D. none of these

Answer 1

The correct option is A $\frac{k}{\sqrt{2}}$

$\cos {25}^{\circ }+\sin {25}^{\circ }=k$

$\frac{1}{\sqrt{2}}\cos {25}^{\circ }+\frac{1}{\sqrt{2}}\sin {25}^{\circ }=\frac{k}{\sqrt{2}}$

$\sin ({45}^{\circ }+{25}^{\circ })=\frac{k}{\sqrt{2}}$

$\sin {70}^{\circ }=\frac{k}{\sqrt{2}}$

$\sin ({90}^{\circ }-{20}^{\circ })=\frac{k}{\sqrt{2}}$

$\cos {20}^{\circ }=\frac{k}{\sqrt{2}}$