Question

If the terminal arm from origin passes through $(4,-7)$ and $\theta$ is the angle made by the arm with the positive $x$-axis, then $\sin \theta =-\frac{m}{\sqrt{65}}$. Here $m$ is

Answer 1

The terminal point passes through $(4,-7)$
The distance from x -axis to the point, $P=7$
The distance from y-axis to the point, $B=4$
Thus, distance from the origin, $H=\sqrt{4^2+7^2}$ $=$ $\sqrt{65}$
The point lies in the third quadrant. Thus, cosine will be positive and sine will be negative.
$\sin \theta =-\frac{P}{H}=-\frac{7}{\sqrt{65}}$

Hence, $m=7$