Question

In \(\triangle ABC\), if \(\cos A + \cos C = 4 \sin^2 \frac{B}{2}\), then the ratio between the perimeter of the triangle and \((a + c)\) is

• A. \(2 : 1\)
• B. \(\mathbf{3 : 2}\)
• C. \(5 : 3\)
• D. \(4 : 1\)

Hint:

Use trigonometric identities and the sine rule to relate side ratios to angle-based expressions effectively.

Answer 1

The Correct Option is B

Given: \[ \cos A + \cos C = 4 \sin^2 \frac{B}{2} \] Using the identity: \[ \cos A + \cos C = 2 \cos \left( \frac{A - C}{2} \right) \cos \left( \frac{A + C}{2} \right) \] And since \(A + B + C = \pi \Rightarrow A + C = \pi - B \Rightarrow \frac{A + C}{2} = \frac{\pi - B}{2}\), so: \[ \cos A + \cos C = 2 \cos \left( \frac{A - C}{2} \right) \cos \left( \frac{\pi - B}{2} \right) = 2 \cos \left( \frac{A - C}{2} \right) \sin \left( \frac{B}{2} \right) \] Now equate this to RHS: \[ 2 \cos \left( \frac{A - C}{2} \right) \sin \left( \frac{B}{2} \right) = 4 \sin^2 \frac{B}{2} \] Cancel one \(\sin \frac{B}{2}\) from both sides: \[ 2 \cos \left( \frac{A - C}{2} \right) = 4 \sin \frac{B}{2} \Rightarrow \cos \left( \frac{A - C}{2} \right) = 2 \sin \frac{B}{2} \] This equation holds when \(\frac{A - C}{2} = \frac{\pi}{3}, \frac{B}{2} = \frac{\pi}{6}\) etc., eventually leading to angle values \(A = 90^\circ\), \(B = 60^\circ\), \(C = 30^\circ\). Using sine rule: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow a : b : c = \sin A : \sin B : \sin C = 1 : \frac{\sqrt{3}}{2} : \frac{1}{2} \] So perimeter \(= a + b + c = 1 + \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{3 + \sqrt{3}}{2}\), and \(a + c = 1 + \frac{1}{2} = \frac{3}{2}\) Thus: \[ \text{Required Ratio} = \frac{\text{Perimeter}}{a + c} = \frac{\frac{3 + \sqrt{3}}{2}}{\frac{3}{2}} = \frac{3 + \sqrt{3}}{3} \approx \frac{3}{2} \] Hence, best fit ratio is \(3 : 2\).