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enter image description here

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?

Please help.

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  • $\begingroup$ Can you find a right-angled triangle and use Pythagoras? $\endgroup$ – Mark Bennet Dec 25 '18 at 2:01
  • $\begingroup$ You mean AC=2r,AB=r and $\angle ABC=90^\circ$ ? $\endgroup$ – user3767495 Dec 25 '18 at 2:02
  • $\begingroup$ That's the one. $\endgroup$ – Mark Bennet Dec 25 '18 at 8:10
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The ABC angle is a right angle, so $12\sqrt{3} \tan(30) = AB$ then $\frac{AB\cdot BC}{2}$ and you got it.

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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?

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