Extend it

Geometry Level pending

Triangle $ABC$ is inscribed in a circle with center at $O$ of radius $3$ . Given that $\angle CAB=45^\circ$ and $AB=3$ , find $AC$ correct to three decimal places.

The answer is 5.796.

2 solutions

Peter van der Linden
Oct 21, 2017

Since AB=3, it follows that triangle AOB has the same side lengths so every angle is 60 degree. This gives angle OAC = OCA = 15 degrees, leading to AOC = 150 degrees. Using law of sine in triangle AOC gives AC.

A Former Brilliant Member
Oct 21, 2017

By the extended law of sines, we have

$\dfrac{3}{\sin C}=2R$ $\implies$ $\dfrac{3}{\sin C}=2(3)$ $\implies$ $C=\sin^{-1}\left(\dfrac{3}{6}\right)=30^\circ$

It follows that $B=180-45-30=105^\circ$ .

By extented law of sines again, we have

$\dfrac{b}{\sin 105}=2R$ $\implies$ $\dfrac{b}{\sin 105}=2(3)$ $\implies$ $b=(\sin 105)(6) \approx$ $\boxed{5.796}$

What's the extended law of sine if in might ask? I used the law of sine 2, but only once in triangle AOC.

Peter van der Linden - 3 years, 7 months ago

Extend it

The answer is 5.796.

2 solutions

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