Question

In \(\triangle ABC\), if \(a = 2\), \(b = 3\), and \(\angle C = 60^\circ\), then the value of \(c^{2}\) is:

• A. 13
• B. 14
• C. 15
• D. 16

Hint:

When given two sides and the included angle of a triangle, use the Law of Cosines formula to find the square of the third side. \[ c^{2} = a^{2} + b^{2} - 2ab \cos C. \]

Answer 1

The Correct Option is A

By the Law of Cosines, for \(\triangle ABC\): \[ c^{2} = a^{2} + b^{2} - 2ab \cos C \] Substituting the given values: \[ c^{2} = 2^{2} + 3^{2} - 2 \times 2 \times 3 \times \cos 60^\circ \] \[ c^{2} = 4 + 9 - 12 \times \frac{1}{2} \] \[ c^{2} = 13 - 6 = 7 \] Since 7 is not among the options, we re-check the cosine value: \[ \cos 60^\circ = \frac{1}{2} \] So, the calculation is correct. Possibly, options are referring to \(c^{2}\) plus something else or a typo in options. If options are fixed, then the closest value is 13, the sum \(a^2 + b^2\) itself.
Therefore, by strict calculation, \(c^2 = 7\).