Question

If co$\sec ({20}^{\circ }+x)=\sec ({50}^{\circ }+x)$ the value of x is

• A. ${10}^{\circ }$
• B. ${20}^{\circ }$
• C. ${30}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A ${10}^{\circ }$

We know that $cosec\left(A\right)=sec(90-A)$
This means, if ${20}^0+x=A$,
then ${50}^0+x={90}^0-({20}^0+x)$
$=>{50}^0+x={90}^0-{20}^0-x$
$=>2x={20}^0$
$=>x={10}^0$