Circumscribed circle

Question

$\triangle ABC: \angle CAB = 45^o, BC = 9$. I have to find the diameter of the circumscribed circle. I thought about an hour, but I can't think up anything. I would be very grateful if you can help me!

I made a drawing:

Answer 1

$\triangle OBC$ is right angled, and $OB=OC$. You'll be able to find the radius.

Answer 2

Hint: Use that $$\sin(\alpha)=\frac{a}{2R}$$

Answer 3

Let the perpendicular on $BC$ drawn from $O$ meet $BC$ at the point $D.$ Then observe that $$\Delta OBD \cong \Delta OCD.$$ So $BD=CD = \frac 9 2.$ Also observe that $$\angle BOC = 2 \times \angle BAC = 90^{\circ}.$$ So $$\angle OBC = \angle OCB = 45^{\circ}.$$ Now let $OC=r.$ Then we have $r=\frac {\frac 9 2} {\cos 45^{\circ}} = \frac {9} {\sqrt 2}.$

So the diameter of the circle is $2r = 9\sqrt 2.$

Another approach $:$ Since $\angle BOC = 90^{\circ}$ so $\Delta OBC$ is a right angled triangle. Let $r$ be the radius of the circle. Now observe that $OB=OC = r.$ Then by Pythagoras theorem we have $$2r^2 =81 \implies r = \frac {9} {\sqrt 2}.$$ So the diameter of the circle is $2r = 9 \sqrt 2.$