circlesgeometrytriangles
An equilateral triangle ABC is inscribed in a circle of radius $6$ cm. O is the centre of the circle. Find its sides.
What I did:-
I joined OB and OA and by applying trigonometry in triangle OAB, tried finding sides of equilateral triangle. But realized it wasn't a right angled triangle. How to find the sides?
From the Cosine Law,
$$AB^2+AC^2-BC^2=2\cdot AB\cdot AC\cdot \cos A \implies AB = 5 \text{ or } 16$$ The area of $\triangle ABC$ can be written as the following two ways: $$S_{ABC}=\frac{1}{2}\cdot AB\cdot AC\cdot \sin A = \frac{1}{2}\cdot(AB+AC+BC)\cdot r$$ where $r$ is the radius of the incircle of $\triangle ABC$. This will give us $$r=\frac{AB\cdot AC\cdot \sin60^\circ}{AB+AC+BC}$$ By plugging in the values of $AB, AC, BC$ and $\sin A$, we can get $r$. Further, because $BP=BQ,\ CQ=CR,\ AR=AP$, and $$BP+AP=AB\\BQ+CQ=BC\\AR+RC=AC$$ we know that $$BQ = \frac{BA+BC-AC}{2}$$Therefore, the following two cases are:
Case 1: $AB=5$, then $r=\frac{7\sqrt{3}}{6}$ and $BQ=\frac{3}{2}$.
Case 2: $AB = 16$, then $r = 3\sqrt{3}$ and $BQ = 7$.
$\angle OBC = 30^\circ$, Hence $BM=r\cos 30^\circ = \frac{\sqrt3 r}{2}.$
Task given to you, find the length of $BC$.
Best Answer
Drop a perpendicular from the center of the circle to one of the sides and draw the radius to one of its endpoints. This should give you some nice triangles to play with :)