Question

In \(△ABC\), with usual notations, \((a+b+c)(b+c-a)(c+a-b)(a+b-c) = 3b^2c^2\), then \(\angle A =\)

• A. \(60^{\circ}\) or \(120^{\circ}\)
• B. \(30^{\circ}\) or \(150^{\circ}\)
• C. \(45^{\circ}\) or \(135^{\circ}\)
• D. \(30^{\circ}\) or \(90^{\circ}\)

Hint:

The product of the four factors is 16 times area squared, which equals 4 b squared c squared sin squared A.

Answer 1

The Correct Option is A

Step 1: Understand the concept
Rewrite two pairs using \((b+c)^2 - a^2\) and \(a^2 - (b-c)^2\): \((a+b+c)(b+c-a) = (b+c)^2 - a^2\) and \((c+a-b)(a+b-c) = a^2 - (b-c)^2\).

Step 2: Use the cosine rule
\((b+c)^2 - a^2 = 2bc(1 + \cos A)\) and \(a^2 - (b-c)^2 = 2bc(1 - \cos A)\), using \(a^2 = b^2 + c^2 - 2bc\cos A\).

Step 3: Multiply
The product is \(4b^2c^2(1 - \cos^2 A) = 4b^2c^2\sin^2 A\). Setting this equal to \(3b^2c^2\):
\[ \sin^2 A = \frac{3}{4} \Rightarrow \sin A = \frac{\sqrt{3}}{2} \]

Step 4: Find A
Since \(A\) is an angle of a triangle, \(A = 60^\circ\) or \(120^\circ\). Both are possible, so option (A). The other options have \(\sin A = \frac{1}{2}\), \(\frac{1}{\sqrt{2}}\) or \(\frac{1}{2}\) and 1, which do not satisfy \(\sin A = \frac{\sqrt{3}}{2}\).

Final Answer:
Angle A is 60 degrees or 120 degrees. This is option (A). \[ \boxed{\text{(A) }60^\circ \text{ or } 120^\circ} \]