Find the Area of triangle in the semi-circle

In the above figure, O is the centre of the circle.

If $$\angle BCO=30 ^\circ$$ and BC=$$12 \sqrt 3$$, what is the area of triangle ABO?

I worked like OA=OB=OC(radii of the circle).

So, $$\angle OBC=30^\circ,\angle BOC=120^\circ$$

$$\angle AOB=60^\circ,\angle ABO=60^\circ,\angle OAB=60^\circ$$

Triangle AOB comes to be an equilateral triangle.

How Do I find OA?

• You mean AC=2r,AB=r and $\angle ABC=90^\circ$ ? – user3767495 Dec 25 '18 at 2:02
The ABC angle is a right angle, so $$12\sqrt{3} \tan(30) = AB$$ then $$\frac{AB\cdot BC}{2}$$ and you got it.
Let $$AO=x$$.
Thus, $$AB=2x$$ and since $$\measuredangle BCO=30^{\circ},$$ we obtain: $$AB=\frac{1}{2}AC=x.$$ Now, by Pythagoras $$AC^2-AB^2=BC^2$$ or $$(2x)^2-x^2=(12\sqrt3)^2,$$ which gives $$x=12.$$ Can you end it now?