Circular measure

Question

Hi everyone,

This is a question from a June 1984 cambridge past paper. I'm getting stuck with the part (c) and the 'hence show...'

Please someone can help, I'd be very grateful.

Answer 1

$OA=OD$ and so $\angle ADO=\theta$, so $\angle AOD=\pi-\theta$. So, area of triangle $AOD$ is $\frac{1}{2}\sin(\pi-2\theta)r\times r=\frac{1}{2}\sin(2\theta)r^2$.

Answer 2

part (a)

$\triangle AOD$ is isosceles triangle with similar sides $= r$ (also called as legs), let $OO'$ be the altitude, since altitude bisects the base,

$l(AD)=2\sqrt{(\text{leg})^2-(\text{altitude})^2}=2\sqrt{r^2-r^2\sin^2 \theta}=2r\cos \theta$

part (b)

Area of the sector formed by $AB, AD$ and arc $BCD$ with center at $A$ is

(fraction of the angle forming the sector) $\times \pi \times (\text {radius})^2$ (why?)

$$\frac{2\theta}{2\pi}\times \pi \times (2r\cos\theta)^2 = 4 r^2 \theta\cos^2 \theta$$