0
$\begingroup$

$\triangle ABC: \angle CAB = 45^o, BC = 9$. 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:

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you! I got it. I appreciate your help. $\endgroup$ – Nikol Dimitrova Apr 15 at 17:30
  • $\begingroup$ You're welcome$\quad$ $\endgroup$ – HAMIDINE SOUMARE Apr 15 at 17:31
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I am in 8th grade and I don't know sin. $\endgroup$ – Nikol Dimitrova Apr 15 at 17:22
0
$\begingroup$

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 $2r = 9 \sqrt 2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.