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: Understanding the Concept
This problem provides two relationships between the sides of a triangle ABC. We need to manipulate these equations, likely using standard triangle formulas like the Cosine Rule or Sine Rule, to determine one of the angles.
Step 2: Key Formula or Approach
The Cosine Rule relates the sides of a triangle to the cosine of one of its angles. For angle C, it is: \[ \cos C = \frac{a^2+b^2-c^2}{2ab} \] We will manipulate the given equations to isolate an expression that matches the form of the Cosine Rule.
Step 3: Detailed Explanation
Let's analyze the two given equations. Equation 1: $c^2 - a^2 = b(\sqrt{3}c - b)$ \[ c^2 - a^2 = \sqrt{3}bc - b^2 \] \[ b^2 + c^2 - a^2 = \sqrt{3}bc \] From the Cosine Rule for angle A, we have $\cos A = \frac{b^2+c^2-a^2}{2bc}$. Substituting our result: \[ \cos A = \frac{\sqrt{3}bc}{2bc} = \frac{\sqrt{3}}{2} \] This implies that angle $A = 30^\circ$. Equation 2: $b^2 - a^2 = c(c-a)$ \[ b^2 - a^2 = c^2 - ac \] \[ a^2 + c^2 - b^2 = ac \] From the Cosine Rule for angle B, we have $\cos B = \frac{a^2+c^2-b^2}{2ac}$. Substituting our result: \[ \cos B = \frac{ac}{2ac} = \frac{1}{2} \] This implies that angle $B = 60^\circ$. We need to find the angle $\angle ACB$, which is angle C. The sum of angles in a triangle is $180^\circ$. \[ A + B + C = 180^\circ \] \[ 30^\circ + 60^\circ + C = 180^\circ \] \[ 90^\circ + C = 180^\circ \] \[ C = 90^\circ \] Step 4: Final Answer
From the given relations, we found angle $A=30^\circ$ and angle $B=60^\circ$. Therefore, angle $C$ must be $90^\circ$.