# Why is $\theta=\beta/2?$ - Kinematics [closed]

An aircraft is flying horizontally in a circle of radius $$b$$ with constant speed $$u$$ at an altitude $$h$$. A radar tracking unit is located at $$C$$. Write expressions for the components of the velocity of the aircraft in the spherical coordinates of the radar station for a given position $$\beta.$$

At some point in the solution to this problem they state that $$\theta=\beta/2$$. I don't see how that is evident from the figure. Can anyone show me how this is deduced?

I also wonder why

$$\sin\phi=\frac{h}{R}\implies\dot{\phi}=-\frac{h\dot{R}}{rR}.$$

If I differentiate both sides I get

$$\dot{\phi}\cos{\phi}=-\frac{h\dot{R}}{R^2}\implies\dot{\phi}=-\frac{h\dot{R}}{R^2\cos{\phi}}.$$

What am I missing?

• To address why $\theta = \frac {\beta}2$, see the diagram from top such that O is the centre of a circle, and C and the aircraft are points on it. You can then deduce the relation by marking the angles there and using angle sum property of a triangle. Jul 18, 2019 at 16:24

Along the $$x-$$direction, the airplane moves a distance of $$b \sin \beta = 2 b \sin \frac{\beta}{2} \cos \frac{\beta}{2}$$. Along the $$y-$$ direction, the distance moved is $$b(1-\cos \beta) = 2 b \sin^2 \frac{\beta}{2}$$. Now $$\tan \theta = \frac{2 b \sin^2 \frac{\beta}{2}}{2 b \sin \frac{\beta}{2} \cos \frac{\beta}{2}}$$ (ratio of distance moved in $$y-$$ direction to distance moved in $$x-$$direction) which is equal to $$\tan \frac{\beta}{2}$$. Thus, $$\theta = \frac{\beta}{2}$$.
$$\cos \phi = \frac{r}{R}$$ to get rid of $$\phi$$.
• Carrying O-plane-z=h triangle to the z=0 plane would be easier. Then, by the isosceles, $2\times(90-\theta)+\beta=180$. Sep 24, 2021 at 22:06
Looking down the $$z$$ axis toward $$C$$, you will see that $$r$$ is the base of an isosceles triangle whose other two sides have length $$b$$. Of the two equal angles in this triangle, one is $$90-\theta$$ and the other is the sum of $$\theta$$ and $$90 - \beta$$. Equating the two angles: $$90 - \theta = \theta + 90 - \beta$$, which leads to $$\beta=2\theta$$.
To see how one angle is the sum of $$\theta$$ and $$90-\beta$$, it may help to draw a horizontal (parallel to the $$x$$ axis) line through the plane's position.