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: Simplify the expression
\[ \frac{2\sin A - \sin 2A}{2\sin A + \sin 2A} = \frac{2\sin A(1 - \cos A)}{2\sin A(1 + \cos A)} = \frac{1 - \cos A}{1 + \cos A} = \tan^2\frac{A}{2} \]

Step 2: Half angle formula
In a triangle, \(\tan\frac{A}{2} = \dfrac{(s-b)(s-c)}{\Delta}\), where \(s = \dfrac{a+b+c}{2}\).

Step 3: Compute s
\(a = 2\), \(b = \frac{7}{2}\), \(c = \frac{5}{2}\), so \(s = \frac{8}{2} = 4\). Then \(s - b = \frac{1}{2}\) and \(s - c = \frac{3}{2}\), so \((s-b)(s-c) = \frac{3}{4}\).

Step 4: Result
\(\tan\frac{A}{2} = \dfrac{3}{4\Delta}\), so the expression equals \(\left(\dfrac{3}{4\Delta}\right)^2\). Option (B). Option (A) is \(\tan\frac{A}{2}\) without the square, and the options with 45 come from using \(s(s-a)\) wrongly.

Final Answer:
The expression is tan squared of A/2, equal to (3/(4 Delta)) squared. This is option (B). \[ \boxed{\text{(B) }\left(\frac{3}{4\Delta}\right)^2} \]