Question

In $△ABC$, angle A is ${120}^{\circ }$, BC + CA = 20, and AB + BC = 21. Find the length of the side BC.

Answer 1

Let $AB=c,BC=a,$ and $CA=b$

$\Rightarrow a+b=20,c+a=21$

$\Rightarrow b=20-a,c=21-a$

We have $\cos A={120}^{\circ }$

Using cosine rule,$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos {120}^{\circ }=\frac{{(20-a)}^2+{(21-a)}^2-a^2}{2(20-a)(21-a)}$

$\Rightarrow \cos ({180}^{\circ }-{60}^{\circ })=\frac{400+a^2-40a+441+a^2-42a-a^2}{2(420-20a-21a+a^2)}$

$\Rightarrow -\cos {60}^{\circ }=\frac{841+a^2-82a}{2(420-20a-21a+a^2)}$

$\Rightarrow \frac{-1}{2}=\frac{841+a^2-82a}{2(420-20a-21a+a^2)}$

$\Rightarrow -1(420-20a-21a+a^2)=841+a^2-82a$

$\Rightarrow -420+41a-a^2-841-a^2+82a=0$

$\Rightarrow -2a^2+123a-1261=0$

$\Rightarrow 2a^2-123a+1261=0$ which is quadratic in $a$

$a=\frac{123\pm \sqrt{{123}^2-4\times 2\times 1261}}{4\times 2}$

$=\frac{123\pm \sqrt{15129-10088}}{8}$

$=\frac{123\pm 71}{8}=\frac{123+71}{8},\frac{123-71}{8}=24.25$units or $6.5$units

$\therefore$ the length of $BC=24.25$units or $6.5$units