Hi :) What properties of triangles should I use to solve this problem? I can work out the size of the large triangle and all of its angles, but how do I apply this to the circle? What am I missing?
This problem is from the first chapter of a trig book, so the methods needed can't be that advanced. Just need some pointers. Thanks!
$BC=1.78cm$
$BD=0.89cm$ ,as the triangle is isosceles.
$\tan{\angle{ABD}}=\tan{70^\circ}=\dfrac{AD}{BD}=\dfrac{AD}{0.89}$
$AD=0.89\tan{70^\circ}$
$AO=AD-r=0.89\tan{70^\circ}-r$
$\sin{\angle{OAE}}=\sin{20^\circ}=\dfrac{OE}{AO}=\dfrac{r}{0.89\tan{70^\circ}-r}$
$0.89\sin{20^\circ}\tan{70^\circ}-r\sin{20^\circ}=r$
$0.89\cos{70^\circ}\tan{70^\circ}=r\left(1+\sin{20^\circ}\right)$
$0.89\sin{70^\circ}=r\left(1+\sin{20^\circ}\right)$
$r=\dfrac{0.89\sin{70^\circ}}{1+\sin{20^\circ}}\approx 0.62cm$