Chapter 13, Problem 18MCQ

To determine

To find:Solve $\Delta ABC$ .

Answer to Problem 18MCQ

  $\begin{array}{l}C={45}^0\\ B={85}^0\\ A={50}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}a=18.6\\ b=24.2\\ c=16.4\end{array}$

Given

  $\begin{array}{l}a=18.6\\ b=24.2\\ c=16.4\end{array}$

Use law of cosines

We get

  $\cos \left(A\right)=\frac{b^2+c^2-a^2}{2bc}$

On substituting values

We get

  $\begin{array}{l}\Rightarrow \cos \left(A\right)=\frac{{\left(24.2\right)}^2+{\left(16.4\right)}^2-{\left(18.6\right)}^2}{2\left(24.2\right)\left(16.4\right)}\\ \Rightarrow \cos \left(A\right)=\frac{585.64+268.96-345.96}{793.76}\\ \Rightarrow \cos \left(A\right)=\frac{508.64}{793.76}\\ \Rightarrow \cos \left(A\right)=0.6408\\ \Rightarrow A={\cos }^{-1}\left(0.6408\right)\\ \Rightarrow A={50}^0\end{array}$

Use law of sines

We get

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin B}{24.2}=\frac{\sin \left({50}^0\right)}{18.6}\\ \Rightarrow \frac{\sin B}{24.2}=\frac{\left(0.766\right)}{18.6}\\ \Rightarrow \sin B=\frac{24.2\left(0.766\right)}{18.6}\\ \Rightarrow \sin B=\frac{18.5372}{18.6}\\ \Rightarrow \sin B=0.9966\\ \Rightarrow B={\sin }^{-1}\left(0.9966\right)\\ \Rightarrow B={85}^0\end{array}$

We know

Sum of angles of triangle ${180}^0$

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {50}^0+{85}^0+C={180}^0\\ \Rightarrow {135}^0+C={180}^0\\ \Rightarrow C={180}^0-{135}^0\\ \Rightarrow C={45}^0\end{array}$

Therefore,

We got

  $\begin{array}{l}C={45}^0\\ B={85}^0\\ A={50}^0\end{array}$