Chapter 12.5, Problem 25PPS

To determine

To find: What are the angles formed between each pair of sides.

Answer to Problem 25PPS

  $\begin{array}{l}A={81}^0\\ B={63}^0\\ C={36}^0\end{array}$

Explanation of Solution

Given information:

Let

  $\begin{array}{l}a=2\\ b=1.8\\ c=1.2\end{array}$

Using law of Cosines we get

  $a^2=b^2+c^2-2bc\cos A$

On substituting values

We get

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ \Rightarrow 2^2={\left(1.8\right)}^2+{\left(1.2\right)}^2-2\left(1.8\right)\left(1.2\right)\cos A\\ \Rightarrow 4=3.24+1.44-4.32\cos A\\ \Rightarrow 4=4.68-4.32\cos A\\ \Rightarrow 4.32\cos A=4.68-4\\ \Rightarrow 4.32\cos A=0.68\\ \Rightarrow \cos A=\frac{0.68}{4.32}\\ \Rightarrow \cos A=0.1574\\ \Rightarrow A={\cos }^{-1}\left(0.1574\right)\\ \Rightarrow A={81}^0\end{array}$

Using law of Sines

We get

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {81}^0}{2}=\frac{\sin B}{1.8}\\ \Rightarrow \sin B=\frac{1.8\sin {81}^0}{2}\\ \Rightarrow \sin B=\frac{1.8\left(0.9877\right)}{2}\\ \Rightarrow \sin B=\frac{1.77786}{2}\\ \Rightarrow \sin B=0.88893\\ \Rightarrow B={\sin }^{-1}\left(0.88893\right)\\ \Rightarrow B={63}^0\end{array}$

We know

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

Using that

We get

  $A+B+C={180}^0$

On substituting values

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {81}^0+{63}^0+C={180}^0\\ \Rightarrow {144}^0+C={180}^0\\ \Rightarrow C={180}^0-{144}^0\\ \Rightarrow C={36}^0\end{array}$

Therefore,

  $\begin{array}{l}A={81}^0\\ B={63}^0\\ C={36}^0\end{array}$