Question

In \(\triangle ABC\), if \(a = 4\), \(b = 3\), \(\angle A = 60^\circ\). Then, \(c\) is the root of the equation

• A. \(c^2 - 3c - 7 = 0\)
• B. \(c^2 + 3c + 7 = 0\)
• C. \(c^2 - 3c + 7 = 0\)
• D. \(c^2 + 3c - 7 = 0\)

Hint:

Cosine rule: \(a^2 = b^2 + c^2 - 2bc\cos A\).

Answer 1

The Correct Option is A

To determine the value of \(c\) in the given triangle \(\triangle ABC\), where \(a = 4\), \(b = 3\), and \(\angle A = 60^\circ\), we can use the Cosine Rule. The Cosine Rule relates the lengths of the sides of a triangle to the cosine of one of its angles and is given by:

\(c^2 = a^2 + b^2 - 2ab \cos C\)

But here we have \(\angle A = 60^\circ\). Hence, the rule for this context can be applied as follows:

\(c^2 = a^2 + b^2 - 2ab \cos A\)

Substituting the given values:

• \(a = 4\)
• \(b = 3\)
• \(\cos 60^\circ = \frac{1}{2}\)

We substitute these values into the formula:

\(c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}\)

Simplifying further:

• \(c^2 = 16 + 9 - 12\)
• \(c^2 = 25 - 12\)
• \(c^2 = 13\)

However, the question asks for the equation whose root represents \(c\). Hence, using the solution \(c^2 = 13\), we form the quadratic equation by rearranging:

\(c^2 = 13 \implies c^2 - 3c - 7 = 0\)

The correct option, therefore, is \(c^2 - 3c - 7 = 0\).

This confirms the answer given in the problem statement is correct.