Question

In \(\triangle ABC\), with usual notations, if \(A=60^{\circ }\) and \(a=4\), then the maximum possible area (in sq. units) of \(\triangle ABC\) is...

• A. \(4\)
• B. \(4\sqrt{3}\)
• C. \(8\)
• D. \(8\sqrt{3}\)

Hint:

Use the cosine rule with a = 4 and A = 60 degrees to get bc at most 16, then use the area formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Cosine rule: \(a^{2}=b^{2}+c^{2}-2bc\cos A\).
With \(A=60^{\circ}\): \(16=b^{2}+c^{2}-bc\).

Step 2: Bound bc
Since \(b^{2}+c^{2}\ge 2bc\), we get \(16\ge 2bc-bc=bc\).
So \(bc\le 16\), with equality when \(b=c\).

Step 3: Area
\(\text{Area}=\dfrac{1}{2}bc\sin 60^{\circ}=\dfrac{\sqrt{3}}{4}bc\)
Maximum area \(=\dfrac{\sqrt{3}}{4}\times 16=4\sqrt{3}\)

Final Answer:
The maximum area is 4 root 3 square units. \[ \boxed{\text{(B) }4\sqrt{3}} \]