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: Use Heron
The product $(a+b+c)(b+c-a)(c+a-b)(a+b-c) = 16\Delta^2$, where $\Delta$ is the area.

Step 2: Area formula
$\Delta = \frac{1}{2}bc\sin A$, so $16\Delta^2 = 4b^2c^2\sin^2 A$.

Step 3: Equate
$4b^2c^2\sin^2A = 3b^2c^2$ gives $\sin^2 A = \frac{3}{4}$.

Step 4: Both angles
$\sin A = \frac{\sqrt{3}}{2}$ gives $A = 60^\circ$ or $120^\circ$, since sine is positive in the first two quadrants and a triangle angle can be obtuse.

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