A triangle ABC is inscribed in a circle where AB is the diameter of the circle. If BC is equal to the radius of the circle and the angle at B (∠ABC) is x times the angle at A (∠BAC), find the value of x.

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Step 1: Since AB is the diameter of the circle, by the Thales theorem, angle ACB = 90°.

Step 2: Let the radius of the circle be r. Then AB = 2r (diameter), and BC = r.

Step 3: In triangle ABC, sides are:
- AB = 2r
- BC = r
- AC = unknown

Step 4: Using the Pythagorean theorem in triangle ABC (right-angled at C):
AB² = AC² + BC²
(2r)² = AC² + r²
4r² = AC² + r²
AC² = 3r²
AC = r√3

Step 5: Let ∠BAC = θ. Then ∠ABC = xθ.

Step 6: Use the Law of Sines:
AB / sin(∠ACB) = BC / sin(∠BAC) = AC / sin(∠ABC)

We know ∠ACB = 90°, so sin(∠ACB) = 1.

Therefore,
AB / 1 = BC / sin(θ)
2r = r / sin(θ)
=> sin(θ) = 1/2
=> θ = 30°

Step 7: Using Law of Sines again for AC and ∠ABC:
AC / sin(∠ABC) = AB / sin(∠ACB)
r√3 / sin(xθ) = 2r / 1
=> sin(xθ) = (r√3) / (2r) = √3 / 2

Step 8: sin(xθ) = √3 / 2
We know sin(60°) = √3 / 2
So, xθ = 60°

Step 9: Since θ = 30°, x * 30° = 60°
=> x = 60° / 30° = 2

Answer: x = 2

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