Chapter 13.5, Problem 32PPS

To determine

To find: Solve the given triangle.

Answer to Problem 32PPS

  $\begin{array}{l}q=16\\ R={107}^0\\ S={48}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}Q={25}^0\\ r=36.2\\ s=28\end{array}$

Using law of Cosines

We get

  $q^2=r^2+s^2-2rs\cos Q$

On substituting values

We get

  $\begin{array}{l}{q}^2=r^2+s^2-2rs\cos Q\\ \Rightarrow q^2={\left(36.2\right)}^2+{\left(28\right)}^2-2\left(36.2\right)\left(28\right)\cos \left({25}^0\right)\\ \Rightarrow q^2=1310.44+784-2027.2\left(0.906\right)\\ \Rightarrow q^2=2094.44-1836.64\\ \Rightarrow q^2=257.8\\ \Rightarrow q=\sqrt{257.8}\\ \Rightarrow q=16\end{array}$

Using law of sines

We get

  $\frac{\sin R}{r}=\frac{\sin Q}{q}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin R}{r}=\frac{\sin Q}{q}\\ \Rightarrow \frac{\sin R}{36.2}=\frac{\sin {25}^0}{16}\\ \Rightarrow \sin R=\frac{36.2\sin {25}^0}{16}\\ \Rightarrow \sin R=\frac{36.2\left(0.423\right)}{16}\\ \Rightarrow \sin R=\frac{15.3}{16}\\ \Rightarrow \sin R=0.957\\ \Rightarrow R={\sin }^{-1}\left(0.957\right)\\ \Rightarrow R={73}^0\end{array}$

Since R is obtuse $R={180}^0-{73}^0={107}^0$

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

Using that

We get

  $Q+R+S={180}^0$

On substituting values

We get

  $\begin{array}{l}Q+R+S={180}^0\\ \Rightarrow {25}^0+{107}^0+S={180}^0\\ \Rightarrow {132}^0+S={180}^0\\ \Rightarrow S={180}^0-{132}^0\\ \Rightarrow S={48}^0\end{array}$

Therefore,

  $\begin{array}{l}q=16\\ R={107}^0\\ S={48}^0\end{array}$