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 :

Correct — C. AB is a diameter, so ∠ACB = 90° — the angle in a semicircle.

Let the radius be r. Then AB = 2r and BC = r.
In right triangle ACB, sin(∠BAC) = BC⁄AB = r⁄2r = 1⁄2
So ∠BAC = 30°

The angles sum to 180°, so ∠ABC = 180 − 90 − 30 = 60°
x = ∠ABC ⁄ ∠BAC = 60⁄30 = 2 → option (c).

Why the others are wrong

• (a)
  1 — x = 1 would make the two acute angles equal at 45° each, which happens when BC = AC. That needs BC = r√2, not the r the question gives.
• (b)
  2.5 — x = 2.5 puts ∠BAC at 90⁄3.5 ≈ 25.7°, and then BC = 2r sin 25.7° ≈ 0.87r — the chord comes out shorter than the radius.
• (d)
  1.5 — x = 1.5 gives ∠BAC = 90⁄2.5 = 36°, so BC = 2r sin 36° ≈ 1.18r, longer than the radius the question specifies.

Two facts do all the work. Thales' theorem says an angle subtended by a diameter at a point on the circle is a right angle, so ∠ACB = 90° and AB is the hypotenuse.

That lets you read BC⁄AB as sin(∠BAC). And because the right angle uses 90° of the 180°, the two acute angles sum to 90° — which means ∠BAC = 90°⁄(1 + x) and one angle always fixes the other.

Key facts

• Thales' theorem: a triangle drawn on a diameter is right-angled at the third vertex.
• A chord equal to the radius subtends 60° at the centre and 30° at the circumference.
• In a right triangle the two acute angles sum to 90°.
• sin 30° = 1⁄2, which is the ratio r : 2r appearing here.

Common traps

• Taking BC = radius to mean the triangle is isosceles — AB is a diameter, so AB = 2r.
• Attaching the ratio to the wrong angle: BC faces ∠BAC, so BC⁄AB is sin(∠BAC).
• Finding ∠ABC = 60° and answering 60 instead of the ratio x the question asks for.

No directly related past PYQ was found.