Question

If ABC is a triangle of area \(\Delta\) with \(a = 2,b = \frac{7}{2},c = \frac{5}{2}\), where \(a,b,c\) are the lengths of the sides of the triangle opposite to angles A, B and C respectively, then \(\frac{2sinA-sin2A}{2sinA+sin2A}\) is equal to...

• A. \(\frac{3}{4\Delta }\)
• B. \((\frac{3}{4\Delta })^2\)
• C. \(\frac{45}{4\Delta }\)
• D. \((\frac{45}{4\Delta })^2\)

Hint:

The expression equals tan squared of A/2, and tan(A/2) = (s-b)(s-c)/area.

Answer 1

The Correct Option is B

Step 1: Half angle reduction
Divide numerator and denominator by $2\sin A$: the ratio is $\dfrac{1-\cos A}{1+\cos A}$.

Step 2: Use the cosine rule
$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{\frac{49}{4} + \frac{25}{4} - 4}{2 \cdot \frac{35}{4}} = \dfrac{\frac{58}{4}}{\frac{70}{4}} = \dfrac{29}{35}$.

Step 3: Evaluate
$\dfrac{1 - \frac{29}{35}}{1 + \frac{29}{35}} = \dfrac{6}{64} = \dfrac{3}{32}$.

Step 4: Compare with options
Heron: $\Delta^2 = 4\cdot2\cdot\frac{1}{2}\cdot\frac{3}{2} = 6$. Then $\left(\frac{3}{4\Delta}\right)^2 = \frac{9}{16 \cdot 6} = \frac{3}{32}$. This agrees, so option (B) is correct.

Final Answer:
The value is 3/32, which equals (3/(4 Delta)) squared. This is option (B). \[ \boxed{\text{(B) }\left(\frac{3}{4\Delta}\right)^2} \]