Chapter 12.9, Problem 3MRE

To determine

Find the length $\angle C$ of triangle ABC to nearest tenth of degree such that $a=6,b=4\text{ and }c=9$ .

Answer to Problem 3MRE

The value of $\angle C$ is equal to $127.2°$ .

Explanation of Solution

Given information: The given triangle ABC has $a=6,b=4\text{ and }c=9$ .

Calculation: By using the law of cosines $\cos C=\frac{a^2+b^2-c^2}{2ab}$ the value of $\angle C$ is evaluated as,

  $\begin{array}{c}\cos C=\frac{{\left(6\right)}^2+{\left(4\right)}^2-{\left(9\right)}^2}{2\left(6\right)\left(4\right)}\mathrm{....}\left(\begin{array}{l}\text{Plugging }a=6,\\ b=4\text{ and }c=9\end{array}\right)\\ ={\cos }^{-1}\left(\frac{{\left(6\right)}^2+{\left(4\right)}^2-{\left(9\right)}^2}{2\left(6\right)\left(4\right)}\right)\\ ={\cos }^{-1}\left(-\frac{29}{48}\right)\\ =127.2°\end{array}$

Hence the value of $\angle C$ is equal to $127.2°$ .