# Triangle inscribed within a unit circle

A triangle $ABC$ is inscribed within the unit circle. Let $x$ be the measure of the angle $C$. Express the length of $AB$ in terms of $x$.

A.) $2\sin x$

B.) $\cos x + \sin x - 1$

C.) $\sqrt{2}(1 - \cos 2x)$

D.) $\sqrt{2}(1 - \sin x)$

I am unsure how to illustrate the circle and calculate the length $x$.

Thank you. `

• Please read this tutorial about how to typeset mathematics on this site. – N. F. Taussig Jun 18 '17 at 8:55

$$\dfrac{AB}{\sin x}=2R$$ where $R$ is the circum-radius $=1$

• This solution is based on the Law of Sines. – N. F. Taussig Jun 18 '17 at 9:11
• @N.F.Taussig, Thanks for enriching the answer – lab bhattacharjee Jun 18 '17 at 9:12

The angle $AOB=2x$, since it intercepts the same arc than the angle $CAB=x$.

thus by the well-known formula

$$AB^2=OA^2+OB^2$$ $$-2OA.OB.\cos (2x)$$ with

$$OA=OB=radius=1$$

hence

$$AB^2=2 (1-\cos (2x))$$

$$=4\sin^2 (x)$$ and $$\boxed {AB=2\sin (x) }$$