Question

sin x=0.6, x lies in quadrant two.

Answer 1

$\qquad\qquad\huge\underline{{\sf Answer}}$

Here we go ~

$\qquad \tt \dashrightarrow \: \sin(x) = 0.6$

$\qquad \tt \dashrightarrow \: \sin(x) = \dfrac{6}{10}$

$\qquad \tt \dashrightarrow \: \sin(x) = \dfrac{3}{5}$

$\qquad \tt \dashrightarrow \:x = \sin {}^{ - 1} ( \dfrac{3}{5} )$

$\qquad \tt \dashrightarrow \:x = 37 \degree$

Now, since we have to find the value of x in quadrant 2, we have to subtract the given value from 180°

that is :

$\qquad \tt \dashrightarrow \:x = 180 - 37$

$\qquad \tt \dashrightarrow \:x = 143 \degree$