Question

In \( \triangle ABC \), if \( (a+c)^2 = b^2 + 3ca \), then \( \frac{a+c}{2R} \) is:

• A. \( \frac{\sqrt{3}}{2} \)
• B. \( \sqrt{3} \cos \left(\frac{A - C}{2} \right) \)
• C. \( \cos \left(\frac{A - C}{2} \right) \)
• D. \( \sin \left(\frac{A - C}{2} \right) \)

Hint:

Use cosine rule transformations to express sides in terms of angles for trigonometric equation-based problems.

Answer 1

The Correct Option is B

We are given a triangle \( \triangle ABC \) with the condition: \[ (a + c)^2 = b^2 + 3ac \] We need to find \( \frac{a + c}{2R} \). Step 1: Expand the Given Equation Expanding both sides, \[ a^2 + c^2 + 2ac = b^2 + 3ac \] Rearranging, \[ a^2 + c^2 - b^2 = ac \] Using the cosine rule, \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \] From the earlier equation, \[ a^2 + c^2 - b^2 = ac \] Thus, \[ \cos B = \frac{ac}{2ac} = \frac{1}{2} \] So, \[ B = 60^\circ \] Step 2: Find \( \frac{a + c}{2R} \) By the sine rule, \[ \frac{a}{\sin A} = 2R \quad \text{and} \quad \frac{c}{\sin C} = 2R \] Now, \[ a + c = 2R \sin A + 2R \sin C = 2R (\sin A + \sin C) \] From the sine addition identity: \[ \sin A + \sin C = 2 \sin \left(\frac{A + C}{2} \right) \cos \left(\frac{A - C}{2} \right) \] Since \( A + C = 180^\circ - B = 120^\circ \), \[ \sin \left(\frac{A + C}{2} \right) = \sin 60^\circ = \frac{\sqrt{3}}{2} \] Thus, \[ \frac{a + c}{2R} = 2 \sin \left(\frac{A + C}{2} \right) \cos \left(\frac{A - C}{2} \right) \] \[ = 2 \times \frac{\sqrt{3}}{2} \times \cos \left(\frac{A - C}{2} \right) \] \[ = \sqrt{3} \cos \left(\frac{A - C}{2} \right) \] Step 3: Final Answer 

\[Correct Answer: (2) \ \sqrt{3} \cos \left(\frac{A - C}{2} \right)\]

Answer 2

In \(\triangle ABC\), we are given the equation \((a+c)^2 = b^2 + 3ca\). Our goal is to find \(\frac{a+c}{2R}\), where \(R\) is the circumradius of the triangle. The expression for the sine rule and circumradius is: \[ R = \frac{a}{2 \sin A} = \frac{b}{2 \sin B} = \frac{c}{2 \sin C} \] The given equation can be expanded and simplified as: \[ a^2 + 2ac + c^2 = b^2 + 3ac \] \[ a^2 + c^2 - b^2 = ac \] Using the cosine rule, \(b^2 = a^2 + c^2 - 2ac \cos B\), substitute to get: \[ ac = 2ac \cos B \] If we divide both sides by \(ac\) (assuming \(a, c \neq 0\)), we have: \[ 1 = 2\cos B \text{ or } \cos B = \frac{1}{2} \] Therefore, \(\angle B = 60^\circ\). So, \(\angle A + \angle C = 120^\circ\). For the bisected angle differences, we use: \[ \cos\left(\frac{A-C}{2}\right) = \sqrt{\frac{1 + \cos(A-C)}{2}} \] The angle \(A-C\) can be found using \(\cos(A-C) = \cos A \cos C + \sin A \sin C \), and knowing \(A+C=120^\circ\). However, simplifying instead through given options, recognize that: \[ \frac{a+c}{2R} = \frac{\sin(A+C)}{2\sin A \sin C} = \frac{\sqrt{3}}{2} \cdot \cos\left(\frac{A-C}{2}\right) \] Thus, the correct option is: \(\sqrt{3} \cos \left(\frac{A - C}{2} \right)\).