Question

A triangle has sides a = 2cm, b = 3cm and ÐC = 60o find the length of side c.

Answer 1

Answer on Question #53687 – Math - Trigonometry

A triangle has sides $a = 2\mathrm{cm}$, $b = 3\mathrm{cm}$ and $C = 60{}^{\circ}$. Find the length of side $c$.

Solution

The Cosine formula is the following:

$$a^{2} - 2ab \cos \gamma + b^{2} = c^{2},$$

where $\gamma$ is the angle against side $c$. Then

$$\begin{array}{l} c = \sqrt{a^{2} - 2ab \cos \gamma + b^{2}} = \sqrt{2^{2} - 2 \cdot 2 \cdot 3 \cdot \cos 60{}^{\circ} + 3^{2}} = \sqrt{4 - 12 \cdot \frac{1}{2} + 9} = \sqrt{4 - 6 + 9} \\ = \sqrt{7} \end{array}$$

Thus, the length of side $c$ is $\sqrt{7}$.

Answer: $\sqrt{7}$.

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