$ABCD$ is a cyclic quadrilateral with diagonal $AC$.

Question

$ABCD$ is a cyclic quadrilateral with diagonal $AC$. Find the area of $\triangle ABC$ where $\angle ABC= 84$, $\angle BAC=36$, $\angle ACB=60$.

I couldn't see any way to get the area. Is there any?

Answer 1

it's not difficult. As one of the angle of $\angle BAD$ and $\angle BCD$ should be greater than or equal to 90 degree (as sum of them must be 180) and sum of the $\angle CAD + \angle ACD = 84$, we can conclude that $\angle CAD$ must be equal 54 and $\angle ACD$ must be equal to 30. So, We have $\angle BAD$ and $\angle BCD$ are equal to 90. Hence, BD is equal to $2r$.

Then using sine rule we can find the size of the BC (as we know all degree of $\triangle BCD$). And using size of BC we can find all value of the other sides of $\triangle ABC$ using sine rule. You can find sine rule as the following:

$\sin(\alpha)/a = \sin(\beta)/b = \sin(\theta)/c$