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:

score 2 · Accepted Answer · answered Apr 15 '19 at 17:17

2

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

answered Apr 15 '19 at 17:17

DINEDINE

6,081

Thank you! I got it. I appreciate your help. – Math Student Apr 15 '19 at 17:30
You're welcome$\quad$ – DINEDINE Apr 15 '19 at 17:31

score 1 · Answer 2 · answered Apr 15 '19 at 17:17

1

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

answered Apr 15 '19 at 17:17

Dr. Sonnhard Graubner

95,283

I am in 8th grade and I don't know sin. – Math Student Apr 15 '19 at 17:22

little o · Answer 3 · 2019-04-15T17:44:42.083

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 $2r = 9 \sqrt 2.$

Circumscribed circle

3 Answers3