Chapter 13, Problem 22PT

To determine

To find: Solve $\Delta ABC$ .

Answer to Problem 22PT

  $\begin{array}{l}Z={27}^0\\ Y={48}^0\\ x=19.4\end{array}$

We should solve beginning with Law of cosines.

Explanation of Solution

Given information:

  $\begin{array}{l}y=15\\ z=9\\ X={105}^0\end{array}$

Given

  $\begin{array}{l}y=15\\ z=9\\ X={105}^0\end{array}$

First, we need to Use law of cosines to solve the given triangle

Use law of cosines

We get

  $\cos \left(X\right)=\frac{y^2+z^2-x^2}{2yz}$

On substituting values

We get

  $\begin{array}{l}\cos \left(X\right)=\frac{y^2+z^2-x^2}{2yz}\\ \Rightarrow \cos \left({105}^0\right)=\frac{{15}^2+9^2-x^2}{2\left(15\right)\left(9\right)}\\ \Rightarrow -0.259=\frac{225+81-x^2}{270}\\ \Rightarrow 225+81-x^2=-0.259\left(270\right)\\ \Rightarrow 306-x^2=-70\\ \Rightarrow x^2=306+70\\ \Rightarrow x^2=376\\ \Rightarrow x=\sqrt{376}\\ \Rightarrow x=19.4\end{array}$

Use law of sines

We get

  $\begin{array}{l}\frac{\sin X}{x}=\frac{\sin Y}{y}\\ \Rightarrow \frac{\sin {105}^0}{19.4}=\frac{\sin Y}{15}\\ \Rightarrow \sin Y=\frac{15\sin {105}^0}{19.4}\\ \Rightarrow \sin Y=\frac{15\left(0.966\right)}{19.4}\\ \Rightarrow \sin Y=\frac{14.49}{19.4}\\ \Rightarrow \sin Y=0.747\\ \Rightarrow Y={\sin }^{-1}\left(0.747\right)\\ \Rightarrow Y={48}^0\end{array}$

We know

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

  $\begin{array}{l}X+Y+Z={180}^0\\ \Rightarrow {105}^0+{48}^0+Z={180}^0\\ \Rightarrow {153}^0+Z={180}^0\\ \Rightarrow Z={180}^0-{153}^0\\ \Rightarrow Z={27}^0\end{array}$

Therefore,

We got

  $\begin{array}{l}Z={27}^0\\ Y={48}^0\\ x=19.4\end{array}$