Question

In a triangle ABC, if $c^2 - a^2 = b(\sqrt{3}c - b)$ and $b^2 - a^2 = c(c-a)$, then $\angle ACB =$

• A. $30^\circ$
• B. $60^\circ$
• C. $45^\circ$
• D. $90^\circ$

Hint:

When given equations involving the squares of the sides of a triangle ($a^2, b^2, c^2$), immediately think of the Cosine Rule. Rearrange the given equations to match the numerator of the Cosine Rule formula ($a^2+b^2-c^2$, etc.) to quickly find the cosines of the angles.

Answer 1

The Correct Option is D

Step 1: Analyze Equation 1 to find Angle A: \[ c^2 - a^2 = \sqrt{3}bc - b^2 \] \[ b^2 + c^2 - a^2 = \sqrt{3}bc \] Using Cosine Rule: \( \cos A = \frac{b^2+c^2-a^2}{2bc} \). \[ \cos A = \frac{\sqrt{3}bc}{2bc} = \frac{\sqrt{3}}{2} \] \[ \angle A = 30^\circ \]
Step 2: Analyze Equation 2 to find Angle B: \[ b^2 - a^2 = c^2 - ac \] \[ a^2 + c^2 - b^2 = ac \] Using Cosine Rule: \( \cos B = \frac{a^2+c^2-b^2}{2ac} \). \[ \cos B = \frac{ac}{2ac} = \frac{1}{2} \] \[ \angle B = 60^\circ \]
Step 3: Find Angle C (ACB): Sum of angles in a triangle is \( 180^\circ \). \[ \angle C = 180^\circ - (A+B) = 180^\circ - (30^\circ + 60^\circ) \] \[ \angle C = 90^\circ \]