 # Find the measures of the angles of the triangle whose vertices are A = (-1,0), B = (2,1), C = (1,-2). opositor5t 2022-08-08 Answered
Find the measures of the angles of the triangle whose vertices are A = (-1,0), B = (2,1), C = (1,-2).
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$\overline{AB}=\sqrt{\left(-1-2{\right)}^{2}+\left(0-1{\right)}^{2}}=\sqrt{9+1}=\sqrt{10}$
$\overline{BC}=\sqrt{\left(1-2{\right)}^{2}+\left(-2-1{\right)}^{2}}=\sqrt{1+9}=\sqrt{10}$
$\overline{AC}=\sqrt{\left(-1-1{\right)}^{2}+\left(0+2{\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$
we will use
${a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}\left(\theta \right)$ where $\theta$ is angle A
we have:
$\mathrm{cos}\left(\mathrm{\angle }ABC\right)=\frac{\overline{A{B}^{2}}+\overline{B{C}^{2}}-\overline{A{C}^{2}}}{2\ast \overline{B{C}^{2}}\ast \overline{A{B}^{2}}}=\frac{10+10-8}{2\ast \sqrt{10}\ast \sqrt{10}}=\frac{12}{20}=\frac{3}{5}$
$\mathrm{\angle }ABC={\mathrm{cos}}^{-1}\left(\frac{3}{5}\right)={53.13}^{0}$
$\mathrm{cos}\left(\mathrm{\angle }ACB\right)=\frac{\left(BC{\right)}^{2}+\left(AC{\right)}^{2}-\left(AB{\right)}^{2}}{2\ast BC\ast AC}=\frac{10+8-10}{2\ast \sqrt{10}\ast \sqrt{8}}=\frac{8}{2\sqrt{80}}=\frac{4}{4\sqrt{5}}=\frac{\sqrt{5}}{5}$
$\mathrm{\angle }BAC={63.43}^{0}$