Chapter 5.6, Problem 16E

Solution

To solve: The triangle if $A={71}^{\circ },a=9.3$ and $b=8.5$ .

The required values are $B={60.94}^{\circ }$, $C={48.06}^{\circ }$ and $c=7$ .

Given information: The given angle and two sides of triangle are $A={71}^{\circ }$

  $a=9.3$ and $b=8.5$ .

Formula used: The law of cosines to find the angles and sides.

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=c^2+a^2-2ca\cos B\end{array}$

Calculation:

Substitute ${71}^{\circ }$ for $\cos A$ , $9.3$ for $a$ and $8.5$ for $b$ in the formula. $a^2=b^2+c^2-2bc\cos A$ .

  $\begin{array}{l}{\left(9.3\right)}^2={\left(8.5\right)}^2+{\left(c\right)}^2-2\left(8.5\right)\left(c\right)\cos {71}^{\circ }\\ 86.46=72.25+c^2-2\left(11.1\right)\left(c\right)\left(0.3255\right)\\ 86.46=72.25+c^2-5.5335c\\ c^2-5.5335c-14.24=0\end{array}$

Further simplify.

  $\begin{array}{l}c=\frac{-\left(-5.5335\right)\pm \sqrt{{\left(-5.5335\right)}^2-4\cdot 1\cdot \left(-14.24\right)}}{2.1}\\ c=7\end{array}$

Substitute $9.3$ for $a$ and $8.5$ for $b$ and $7$ for $c$ in the formula $b^2=c^2+a^2-2ca\cos B$

  $\begin{array}{c}{\left(8.5\right)}^2={\left(7\right)}^2+{\left(9.3\right)}^2-2\left(7\right)\left(9.3\right)\cos B\\ 72.25=49+86.49-130.2\cos B\\ \cos B=\frac{17}{35}\\ B={60.94}^{\circ }\end{array}$

To find the angle $C$ subtract the sum of angles $A$ and $B$ from sum of all three angles of a triangle.

  $\begin{array}{l}C={180}^{\circ }-\left({71}^{\circ }+{60.94}^{\circ }\right)\\ C={48.06}^{\circ }\end{array}$

Thus the required values are $B={60.94}^{\circ }$, $C={48.06}^{\circ }$and $c=7$ .