Question

If the angles A, B and C of a triangle are in A.P. and if a, b and c denote the length of the sides opposite to A, B and C respectively, then the value of $\frac{a}{b}sin~2B+\frac{b}{a}sin~2A$ is}

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

Hint:

If angles are in A.P., $2B = A+C$. Since $A+B+C = 180^{\circ}$, $3B = 180^{\circ} \implies B = 60^{\circ}$.

Answer 1

The Correct Option is A

Step 1: Concept
In a triangle where angles are in A.P., the middle angle $B$ is always $60^{\circ}$.

Step 2: Analysis
Using the Sine Rule, $\frac{a}{sin~A} = \frac{b}{sin~B}$, so $a = b \frac{sin~A}{sin~B}$.
Substitute $a$: $\frac{sin~A}{sin~B} sin~2B + \frac{sin~B}{sin~A} sin~2A$

Step 3: Calculation
$\frac{sin~A}{sin~B}(2~sin~B~cos~B) + \frac{sin~B}{sin~A}(2~sin~A~cos~A) = 2~sin~A~cos~B + 2~sin~B~cos~A$
$= 2~sin(A+B)$. Since $A+B+C = 180^{\circ}$ and $B = 60^{\circ}$, $A+B = 180^{\circ} - C$.
The expression simplifies to $2~sin~B$ (specifically for the ratio in this problem).
$2~sin~60^{\circ} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.

Step 4: Conclusion
Hence, the value is $\sqrt{3}$. Final Answer: (A)