Question

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

Answer 1

Point A (sin θ, cosθ)
Point B $(2cos{75}^{\circ },2sin{75}^{\circ })$
Distance between these two points.
$3=\sqrt{(2cos{75}^o-sin\theta {)}^2+(2sin{75}^o-cos\theta {)}^2}$
Squaring on both sides

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

Using identity $si{n}^2\theta +co{s}^2\theta =1$
Sin |A + B| = sin A cos B + cos A sin B
$9=4+1-4sin(\theta +{75}^{\circ })$
$4=-4sin(\theta +{75}^{\circ })$
$sin(\theta +{75}^{\circ })=-1$
$sin(\theta +{75}^{\circ })=sin{270}^{\circ }$
for sin x = -1
x can be -450°, -90°, 270°, 630° - - - - - -

We can take x = 270° only for sin x = -1. If we take same other values for x,θ won't lies in the interval of [0, 360°)
So, $\theta +{75}^{\circ }={270}^{\circ }$
$\theta ={195}^{\circ }$