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A, B, C are three points on the circumference of a circle and if AB=AC $$5\sqrt{2}$$ BAC = 90°, find the radius.
Given : AB = AC = $$5\sqrt{2}$$ and $$\angle$$BAC = 90°
To find : OB = OC = OA = $$r$$
Solution : SInce, AB = AC, => $$\angle$$ABC = $$\angle$$ACB
In $$\triangle$$ABC,
=> $$\angle$$ABC + $$\angle$$ACB + 90° = 180°
=> $$\angle$$ABC = 45°
Now, in $$\triangle$$OAB
=> $$sin \angle ABO = \frac{OA}{AB}$$
=> $$sin 45^{\circ} = \frac{OA}{5\sqrt{2}}$$
=> $$OA = \frac{5\sqrt{2}}{\sqrt{2}}$$
=> OA = 5 cm
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