Chapter 8, Problem 3CT

To determine

The three remaining parts of the triangle below

  

Answer to Problem 3CT

Solution:

The third side of triangle is $a=15.8834$, and other angles are $B={57.50}^0,C={70.50}^0$ .

Explanation of Solution

Given information:

The triangle

  

To solve the triangle withtwo sides and one angle given,use the law of cosine rule.

Here, $b=17,c=19$, and $A={52}^0$ .

Law of cosine:

  $a^2=b^2+c^2-2bc\cdot \cos A$

By substituting values of $b,c$, and $A$ ,

  $\Rightarrow a^2={17}^2+{19}^2-2\cdot 17\cdot 19\cdot \cos \left(52\right)$

  $\Rightarrow a^2=289+361-646\cdot \left(0.61566\right)$

  $\Rightarrow a^2=252.2827$

  $\Rightarrow a=15.8834$

To find angle $B$ :

  $\cos B=\frac{a^2+c^2-b^2}{2ac}$

  $\cos B=\frac{{\left(15.8834\right)}^2+{\left(19\right)}^2-{\left(17\right)}^2}{2\left(15.8834\right)\left(19\right)}$

  $\cos B=\frac{\left(252.2827\right)+\left(361\right)-\left(289\right)}{603.5692}=\frac{324.2827}{603.5692}$

  $B={\cos }^{-1}\left(\frac{324.2827}{603.5692}\right)\approx {57.50}^0$ .

To find angle $C$ :

  $\cos C=\frac{a^2+b^2-c^2}{2ab}$

  $\cos C=\frac{{\left(15.8834\right)}^2+{\left(17\right)}^2-{\left(19\right)}^2}{2\left(15.8834\right)\left(17\right)}$

  $\cos C=\frac{\left(252.2827\right)+\left(289\right)-\left(361\right)}{540.0356}=\frac{180.2827}{540.0356}$

  $C={\cos }^{-1}\left(\frac{180.2827}{540.0356}\right)\approx {70.50}^0$ .

Thus, the third side of the triangle is $a=15.8834$, and other angles are $B={57.50}^0,C={70.50}^0$ .