A line OA of length $ r$ starts from its initial position OX and traces an angle AOB=$\alpha$ in the anticlockwise direction.It then traces back in the clockwise direction an angle BOC=3 $\theta $(where alpha is greater than 3 $\theta$).L is the foot of the perpendicular form C on OA.
Also $\frac{\sin ^3 \theta}{CL} = \frac{\cos^3 \theta}{OL} = 1$
1) $ \frac{1-r\cos \alpha }{r \sin \alpha} $ is equal to..
answer is $\tan 2\theta $
2)$ \frac{2r\sin \alpha }{1+2r\cos \alpha} $ is equal to..
answer is $\tan 2\theta $
3)$ \frac{2r^2 - 1}{r } $ is equal to ..
answer is $\cos \alpha $
So I have found out that CL is $ r sin( \alpha - 3\theta) $ and OL is $ r cos( \alpha - 3\theta) $ I have tried substituting r with different ways. They don't work out. I have tried constructing a right triangle OBP and solving it. I have tried messing with area.
I have no idea how to title this question. Please help me with that.
