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 

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