Let the point $A$ lie on the exterior of the circle $k(R).$ From $A$ are drawn the tangents $AB$ and $AC$ to $k.$ The triangle $ABC$ is еquilateral. Find the side of $\triangle ABC$.

Answer: $R\sqrt{3}.$

enter image description here I am not sure how to approach the problem. We should use the Intersecting Chords Theorem. Can you give me a hint?

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    $\begingroup$ The triangle $BOC$ is isosceles and the base angles are 30 degrees. $\endgroup$ – Catalin Zara May 30 at 17:15
  • $\begingroup$ Another way: for $ABOC$ the sum of angles is $360^\circ$, but $\angle B=\angle C=90^\circ$, so $\angle BOC=180^\circ -\angle CAB=120^\circ$, then apply the cosine rule for $\triangle BOC$ $\endgroup$ – Alexey Burdin May 30 at 22:28


Join $O$ and $A$. $OA$ bisects $\angle ABC$ and so $\angle OAB=30^\circ$. What is $\tan\angle OAB$?

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    $\begingroup$ Hints should be given in comments first. Don't simply post hints in answers just for votes. $\endgroup$ – SarGe Jun 4 at 4:00
  • $\begingroup$ @Doubtnut The OP specifically asked for a hint, so what’s wrong with posting it as an answer? $\endgroup$ – Tavish Jun 4 at 9:46
  • $\begingroup$ Comments are used for hints. $\endgroup$ – SarGe Jun 4 at 10:40
  • $\begingroup$ @Doubtnut Not necessarily. I see many people post hints as answers. $\endgroup$ – Tavish Jun 4 at 11:27
  • $\begingroup$ It is not necessary that you should copy the same. Read this $\endgroup$ – SarGe Jun 4 at 11:30

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