The area of the triangle $\mathrm{ABC}$ is $10 \sqrt{3} \mathrm{~cm}^2$, angle $\mathrm{B}$ is $60^{\circ}$…
The area of the triangle $\mathrm{ABC}$ is $10 \sqrt{3} \mathrm{~cm}^2$, angle $\mathrm{B}$ is $60^{\circ}$ and its perimeter is $20 \mathrm{~cm}$, then $\ell(\mathrm{AC})=$
10 cm
8 cm
5 cm
7 cm
Solution
$\begin{aligned}
& \text { Area }=\frac{\operatorname{acsin} B}{2}=\Rightarrow 10 \sqrt{3}=\frac{\operatorname{acsin} 60^{\circ}}{2} \\
& 10 \sqrt{3}=\frac{a c \sqrt{3}}{4} \Rightarrow a c=40
\end{aligned}$
Now
$\begin{aligned}
& \cos B=\frac{a^2+c^2-b^2}{2 a c} \\
& \cos 60^{\circ} \times 2 a c=a^2+c^2-b^2 \\
& \therefore \frac{1}{2} \times 2 \times 40=\left(a^2+c^2\right)-2 a c-b^2 \\
& \therefore 40=(20-b)^2-(2 \times 40)-b^2 \quad \ldots[a+b+c=20, \text { given }] \\
& =400+b^2-40 b-80-b^2 \\
& \therefore 40 b=280 \Rightarrow b=7
\end{aligned}$