In triangle $ABC$, side $a=8\text{cm}$, side $b=10\text{cm}$ and the included angle $C=60^{\circ }$. Find the length of side $c$ (correct to two decimal places).

Step-by-step solution

1. Use the cosine rule: $c^2=a^2+b^2-2ab\cos C$.
2. Substitute the given values: $c^2=8^2+10^2-2\times 8\times 10\cos 60^{\circ }$.
3. Compute $8^2+10^2=64+100=164$.
4. Since $\cos 60^{\circ }=0.5$, evaluate $2\times 8\times 10\times 0.5=80$.
5. Thus $c^2=164-80=84$.
6. Take the square root: $c=\sqrt{84}\approx 9.165\text{cm}$.
7. Rounded to two decimal places, $c=9.17\text{cm}$.

Concept tested: Cosine rule to find a side

Published 20 August 2026 · How this solution was prepared and checked