Chapter 12.5, Problem 22PPS

To determine

To find: Solve the given triangle.

Answer to Problem 22PPS

  $\begin{array}{l}K={107}^0\\ H={48}^0\\ J={25}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}h=18\\ j=10\\ k=23\end{array}$

Given all three sides of a triangle.

So, the triangle should be solved by beginning with law of cosines.

Using law of Cosines we get

  $k^2=h^2+j^2-2hj\cos K$

On substituting values

We get

  $\begin{array}{l}{k}^2=h^2+j^2-2hj\cos K\\ \Rightarrow {23}^2={18}^2+{10}^2-2\left(18\right)\left(10\right)\cos K\\ \Rightarrow 529=324+100-360\cos K\\ \Rightarrow 529=424-360\cos K\\ \Rightarrow 360\cos K=424-529\\ \Rightarrow 360\cos K=-105\\ \Rightarrow \cos K=\frac{-105}{360}\\ \Rightarrow \cos K=-0.292\\ \Rightarrow K={\cos }^{-1}\left(-0.292\right)\\ \Rightarrow K={107}^0\end{array}$

Using law of sines

We get

  $\frac{\sin K}{k}=\frac{\sin H}{h}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin K}{k}=\frac{\sin H}{h}\\ \Rightarrow \frac{\sin {107}^0}{23}=\frac{\sin H}{18}\\ \Rightarrow \sin H=\frac{18\sin {107}^0}{23}\\ \Rightarrow \sin H=\frac{18\left(0.9563\right)}{23}\\ \Rightarrow \sin H=\frac{17.21}{23}\\ \Rightarrow \sin H=0.7484\\ \Rightarrow H={\sin }^{-1}\left(0.7484\right)\\ \Rightarrow H={48}^0\end{array}$

We know

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

Using that

We get

  $H+J+K={180}^0$

On substituting values

We get

  $\begin{array}{l}H+J+K={180}^0\\ \Rightarrow {107}^0+{48}^0+J={180}^0\\ \Rightarrow {155}^0+J={180}^0\\ \Rightarrow J={180}^0-{155}^0\\ \Rightarrow J={25}^0\end{array}$

Therefore,

  $\begin{array}{l}K={107}^0\\ H={48}^0\\ J={25}^0\end{array}$