Bunuel wrote:
The triangle ABC is isosceles. What is the measure of the angle ACB?
(1) angle ABC = 2 × angle ACB
(2) BC < AC
Kudos for a correct solution.
Very tempting question to pick wrong answer
Stmt1 - ∠B = 2 * ∠C . Lets take ∠C = x. Then ∠B = 2x.
An isosceles triangle has two equal angles. There are two possible options i.e, ∠A = ∠C or , ∠A = ∠B.
For ∠A = ∠C, sum of the angles in the triangle is x + x + 2x = 4x. Sum of any triangle must be 180°. 4x = 180°. So x = 45°. In this case ∠C = 45°.
For ∠A = ∠B, sum of the angles in the triangle is x + 2x + 2x = 5x. So 5x = 180° and x = 36°. In this case ∠C = 36°. We have two different possible values of the ∠C. Not Sufficient.
Stmt2 - BC < AC. In any triangle, the greater angle is subtended by the greater side. So we can say that ∠B > ∠A. But we do not know about ∠C. Not Sufficient.
1+2, from stmt2 ∠B > ∠A. So we have only one option left from Stmt1 ∠C = ∠A . ----> ∠C = 45° . Sufficient.
ANS C.