A triangle ABC is made on a circle, where AB is diameter of the circle. If BC is equal to the radius of the circle and ∠ABC = x ∠BAC, then x is:

Triangle ABC is inscribed in a circle with AB as the diameter.

BC is equal to the radius of the circle.

∠ABC = x ∠BAC.

∠ABC in a triangle inscribed in a circle where AB is the diameter is a right angle (90°), based on the property of a triangle inscribed in a semicircle.

An equilateral triangle has equal sides and every interior angle $$60^\circ$$​.

From the fig., 

OC = OB , ( radius)

OB = BC (given) 

So, $$\triangle OBC$$ is a equilateral triangle ,

Thus, $$\angle OBC =\angle BCO =\angle BOC = 60^\circ$$ 

Now, $$\angle ACB = 90^\circ$$  ( angle made by diameter AB)

$$\angle ACB = \angle ACO +\angle BCO$$ 

$$\implies \angle ACO = 90 -60 = 30 ^\circ$$ 

Now, AO = CO (radius) 

So, $$\angle ACO = \angle BAC = 30^\circ$$  

Now

  $$\angle ABC = x \angle BAC \\ 60^\circ = x \times 30^\circ \\ x = \frac{60^\circ}{30^\circ} =\bf 2$$