Question

The point A(sin $\theta$, cos$\theta$) is 3 units away from the point B (2 cos ${75}^{\circ }$, 2 sin ${75}^{\circ }$). If $0^{\circ }$ $\le \theta <$ ${360}^{\circ }$, then $\theta$ is __ degree

Answer 1

Point A (sin $\theta$, cos$\theta$)

Point B (2 cos ${75}^{\circ }$, 2 sin ${75}^{\circ }$)

Distance between these two points.

$3=\sqrt{(2cos{75}^{\circ }-sin\theta {)}^2+(2sin{75}^{\circ }-cos\theta {)}^2}$

Squaring on both sides

9 = 4 $co{s}^2{75}^{\circ }$ + $si{n}^2\theta$ - 4 sin$\theta$cos ${75}^{\circ }$ + 4 $si{n}^2{75}^{\circ }$+ $co{s}^2\theta$ - 4 sin ${75}^{\circ }$ . cos$\theta$

9 = 4($si{n}^2{75}^{\circ }$ + $co{s}^2{75}^{\circ }$) + ($si{n}^2\theta$ + $co{s}^2\theta$)- 4 (sin $\theta$cos ${75}^{\circ }$ + cos$\theta$ sin ${75}^{\circ }$)

Using identity $si{n}^2\theta$ + $co{s}^2\theta$ = 1

Sin (A + B) = sin A cos B + cos A sin B

9 = 4 + 1 - 4 sin ($\theta$ + ${75}^{\circ }$)

4 = -4 sin ($\theta$ + ${75}^{\circ }$)

sin ($\theta$ + ${75}^{\circ }$) = -1

sin($\theta$ + ${75}^{\circ }$) = sin${270}^{\circ }$

for sin x = -1

x can be $-{450}^{\circ }$, $-{90}^{\circ }$, ${270}^{\circ }$, ${630}^{\circ }$ - - - - - -

We can take x = ${270}^{\circ }$ only for sin x = -1. If we take same other values for x,$\theta$ won't lie in the interval of [0, ${360}^{\circ }$)

So, $\theta$ + ${75}^{\circ }$ = ${270}^{\circ }$

$\theta$ = ${195}^{\circ }$