Chapter 12.5, Problem 18PPS

To determine

To find: To determine whether each triangle is solved by cosines rule or sine rule in the beginning. Also have to find the other sides and angles.

Answer to Problem 18PPS

We will use cosine rule in the beginning because one angle is given, and two sides are given making this angle.

In triangle RST

So side RT= s= 29

Angle $\angle R$ = ${42}^{\circ }$

And Angle $\angle T$ = ${32}^{\circ }$

Explanation of Solution

Given information:

The length of two sides of triangle RST and one angle.

  RS= t = 16

  ST= r =20

  $\angle$RST= $\angle S$ = ${106}^{\circ }$

Calculation:

We will use cosine rule in the beginning because one angle is given and two sides are given making this angle.

To find the length of side RT= s , we use the cosine formula for any triangle:

  $\begin{array}{l}{s}^2=r^2+t^2-2rt\cos S\\ \Rightarrow s^2={20}^2+{16}^2-2\times 20\times 16\times \cos {106}^{\circ }\\ \Rightarrow s^2=400+256-640\times (-0.2756)\\ \Rightarrow s^2=656+176.40\\ \Rightarrow s^2=832.40\\ \Rightarrow s=\sqrt{\text{101}\text{.08}}\\ \Rightarrow s=28.85\\ \Rightarrow s=29\end{array}$

Also as we know that-

  $\begin{array}{l}\frac{\mathrm{Sin}RST}{s}=\frac{\mathrm{Sin}SRT}{r}\\ \Rightarrow \frac{\mathrm{Sin}S}{s}=\frac{\mathrm{Sin}R}{r}\\ \Rightarrow \mathrm{Sin}R=\frac{r\times \mathrm{Sin}S}{s}\\ \Rightarrow \mathrm{Sin}R=\frac{20\times \mathrm{Sin}{106}^{\circ }}{29}\\ \Rightarrow \mathrm{Sin}R=\frac{20\times 0.9612}{29}\\ \Rightarrow \mathrm{Sin}R=\text{0}\text{.6629}\\ \Rightarrow R={\sin }^{-1}(\text{0}\text{.9177})\\ \Rightarrow R={41.52}^{\circ }\\ \Rightarrow R={42}^{\circ }\\ \end{array}$

Since the sum of the angles of any triangle is ${180}^{\circ }$

So

  $\begin{array}{l}R+S+T={180}^{\circ }\\ \Rightarrow T={180}^{\circ }-(R+S)\\ \Rightarrow T={180}^{\circ }-({42}^{\circ }+{106}^{\circ })\\ \Rightarrow T={180}^{\circ }-{148}^{\circ }\\ \Rightarrow T={32}^{\circ }\\ \end{array}$

In triangle RST

So side RT= s= 29

Angle $\angle R$ = ${42}^{\circ }$

And Angle $\angle T$ = ${32}^{\circ }$