Chapter 6, Problem 27CR

To determine

To calculate:

Solve the triangle using law of cosine.

Answer to Problem 27CR

Angle sof $△ABC$ be $\angle A=45.75°,\angle B=91.24°,c=21.42$

Explanation of Solution

Given information:

  $C=43°,a=22.5,b=31.4$

Calculation:

Let’s,

  $C=43°,a=22.5,b=31.4$

Grah for $C=43°,a=22.5,b=31.4$ ,

For solving the triangle by using law of cosine, need to determine angle opposite to longest side

  $\begin{array}{l}\cos C=\frac{a^2+b^2+-c^2}{2ab}\\ c^2=a^2+b^2-2ab\cos C\end{array}$

Put the values in above equation,

  $\begin{array}{l}{c}^2={\left(22.5\right)}^2+{\left(31.4\right)}^2-2\left(22.5\right)\left(31.4\right)\cos \left(43°\right)\\ c^2=506.25+985.96-1413\left(0.7313\right)\\ c^2=1492.21-1033.33\\ c^2=458.88\\ c=21.42\end{array}$

Similarly,

  $\cos B=\frac{a^2+c^2-b^2}{2ac}$

Put the values in above equation,

  $\begin{array}{l}\cos B=\frac{{\left(22.5\right)}^2+{\left(21.42\right)}^2-{\left(31.4\right)}^2}{2\left(22.5\right)\left(21.42\right)}\\ \cos B=\frac{506.25+458.82-985.96}{963.9}\\ \cos B=\frac{965.07-985.96}{963.9}\\ \cos B=\frac{-20.89}{963.9}\\ \cos B=-0.0217\\ B={\cos }^{-1}\left(-0.0217\right)\\ B=91.24°\end{array}$

Sum of three angle is, $△ABC$

  $\begin{array}{l}\angle A+\angle B+\angle C=180°\\ \angle A+91.25°+43°=180°\\ \angle A=180°-134.25°\\ \angle A=45.75°\end{array}$

Therefore, angle sof $△ABC$ be $\angle A=45.75°,\angle B=91.24°,c=21.42$