Question

In \( \triangle ABC \), if \( A, B, C \) are in arithmetic progression, then \( \frac{c}{a} \sin 2A + \frac{a}{c} \sin 2C = \):

• A. \( \frac{\sqrt{3}}{2} \)
• B. \( \sqrt{3} \)
• C. \( 1 \)
• D. \( \frac{1}{2} \)

Hint:

For triangle problems involving angles in A.P., the middle angle is always \( 60^\circ \). Remember to use the Sine Rule and sum of angles property effectively.

Answer 1

The Correct Option is B

Step 1: Use the property of arithmetic progression and the sum of angles in a triangle.
Since \( A, B, C \) are in arithmetic progression, let the common difference be \( d \). Then \( A = B - d \) and \( C = B + d \). The sum of the angles in a triangle is \( 180^\circ \): \[ A + B + C = 180^\circ \] \[ (B - d) + B + (B + d) = 180^\circ \] \[ 3B = 180^\circ \implies B = 60^\circ \] Thus, \( A + C = 180^\circ - B = 180^\circ - 60^\circ = 120^\circ \).
Step 2: Apply the Sine Rule and trigonometric identities.
Using the Sine Rule, we have \( \frac{a}{\sin A} = \frac{c}{\sin C} \), which implies \( \frac{c}{a} = \frac{\sin C}{\sin A} \) and \( \frac{a}{c} = \frac{\sin A}{\sin C} \).
Substitute these into the expression: \[ \frac{c}{a} \sin 2A + \frac{a}{c} \sin 2C = \frac{\sin C}{\sin A} (2 \sin A \cos A) + \frac{\sin A}{\sin C} (2 \sin C \cos C) \] \[ = 2 \sin C \cos A + 2 \sin A \cos C \] Using the sine addition formula \( \sin(X + Y) = \sin X \cos Y + \cos X \sin Y \): \[ = 2 (\sin C \cos A + \cos C \sin A) = 2 \sin(C + A) \] Since \( A + C = 120^\circ \), we have: \[ 2 \sin(C + A) = 2 \sin(120^\circ) \] We know that \( \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Therefore, \[ 2 \sin(120^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \]