Chapter 12.5, Problem 36HP

To determine

To find: Solve the triangle.

Answer to Problem 36HP

  $\begin{array}{l}H={66}^0\\ G={72}^0\\ F={42}^0\end{array}$

Explanation of Solution

Given information:

Let

  $\begin{array}{l}f=15.2\\ g=21.6\\ h=20.8\end{array}$

Using law of Cosines

We get

  $g^2=h^2+f^2-2hf\cos G$

On substituting values

We get

  $\begin{array}{l}{g}^2=h^2+f^2-2hf\cos G\\ \Rightarrow {\left(21.6\right)}^2={\left(20.8\right)}^2+{\left(15.2\right)}^2-2\left(20.8\right)\left(15.2\right)\cos G\\ \Rightarrow 466.56=432.64+231.04-632.32\cos G\\ \Rightarrow 466.56=663.68-632.32\cos G\\ \Rightarrow 632.32\cos G=663.68-466.56\\ \Rightarrow 632.32\cos G=197.12\\ \Rightarrow \cos G=\frac{197.12}{632.32}\\ \Rightarrow \cos G=0.312\\ \Rightarrow G={\cos }^{-1}\left(0.312\right)\\ \Rightarrow G={72}^0\end{array}$

Using law of sines

We get

  $\frac{\sin H}{h}=\frac{\sin G}{g}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin H}{h}=\frac{\sin G}{g}\\ \Rightarrow \frac{\sin H}{20.8}=\frac{\sin {72}^0}{21.6}\\ \Rightarrow \sin H=\frac{20.8\sin {72}^0}{21.6}\\ \Rightarrow \sin H=\frac{20.8\left(0.951\right)}{21.6}\\ \Rightarrow \sin H=\frac{19.78}{21.6}\\ \Rightarrow \sin H=0.916\\ \Rightarrow H={\sin }^{-1}\left(0.916\right)\\ \Rightarrow H={66}^0\end{array}$

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

Using that

We get

  $F+G+H={180}^0$

On substituting values

We get

  $\begin{array}{l}F+G+H={180}^0\\ \Rightarrow {66}^0+{72}^0+F={180}^0\\ \Rightarrow {138}^0+F={180}^0\\ \Rightarrow F={180}^0-{138}^0\\ \Rightarrow F={42}^0\end{array}$

Therefore,

  $\begin{array}{l}H={66}^0\\ G={72}^0\\ F={42}^0\end{array}$