In triangle ABC, $a=15$, $b=14$, $c=10$. Find $m<B$

Answer & Explanation

yunitsiL

Skilled2021-06-15Added 108 answers

The given case is SSS (three sides) so use the Law of Cosines: ${b}^{2}={a}^{2}+{c}^{2}-2a\mathcal{o}sB$
Solve for B. $2a\mathcal{o}sB={a}^{2}+{c}^{2}-{b}^{2}$ $\mathrm{cos}B=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2}ac$ $m<B=\left({\mathrm{cos}}^{-1}\right)\left(\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\right)$ $m<B=\left({\mathrm{cos}}^{-1}\right)\left(\frac{{15}^{2}+{10}^{2}-{14}^{2}}{2\left(15\right)\left(10\right)}\right)$ $m<B={\mathrm{cos}}^{-1}\left(\frac{129}{300}\right)$
Use a calculator in DEGREE mode: $m\angle B\approx 64.5\xb0$