Question

a triangle has sides a = 6 and b = 2 and angle c = 60°. find the length of side c. the length of side c is (type an exact answer, using radicals as needed.)

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\) is given by \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\).
Here, \(a = 6\), \(b=2\) and \(C = 60^{\circ}\), and \(\cos(60^{\circ})=\frac{1}{2}\).

Step2: Substitute the values into the formula

Substitute \(a = 6\), \(b = 2\) and \(\cos(C)=\frac{1}{2}\) into the formula:
\[

$$\begin{align*} c^{2}&=6^{2}+2^{2}-2\times6\times2\times\cos(60^{\circ})\\ &= 36 + 4-2\times6\times2\times\frac{1}{2}\\ &=40 - 12\\ &= 28 \end{align*}$$

\]

Step3: Solve for \(c\)

Take the square root of both sides to find \(c\). Since \(c\) represents the length of a side of a triangle, we take the positive square root.
\(c=\sqrt{28}=\sqrt{4\times7} = 2\sqrt{7}\)

Answer:

\(2\sqrt{7}\)