Chapter 13.5, Problem 16PPS

To determine

To find: To find the angles of the triangle to the nearest degree.

Answer to Problem 16PPS

In triangle WXY

Angle $\angle W={106}^{\circ }$

Angle $\angle X={39}^{\circ }$

Angle $\angle Y={35}^{\circ }$

Explanation of Solution

Given information:

The length three sides of triangle WXY.

XY= w = 20

WY= x = 13

WX= y= 12

Calculation:

To find the angle W, X and Y we use the cosine formula for any triangle:

  $\begin{array}{l}{x}^2=w^2+y^2-2wy\cos X\\ \Rightarrow {13}^2={20}^2+{12}^2-2\times 20\times 12\times \cos X\\ \Rightarrow 169=400+144-480\mathrm{Cos}X\\ \Rightarrow 169=544-480\mathrm{Cos}X\\ \Rightarrow \mathrm{Cos}X=\frac{544-169}{480}\\ \Rightarrow \mathrm{Cos}X=\frac{373}{480}\\ \Rightarrow \mathrm{Cos}X=0.77708\\ \Rightarrow X={\cos }^{-1}(0.77708)\\ \Rightarrow X={39.08}^{\circ }\\ \Rightarrow X={39}^{\circ }\end{array}$

Also as we know that-

  $\begin{array}{l}{w}^2=x^2+y^2-2wy\cos W\\ \Rightarrow {20}^2={13}^2+{12}^2-2\times 13\times 12\times \cos W\\ \Rightarrow 400=169+144-312\mathrm{Cos}W\\ \Rightarrow 400=313-312\mathrm{Cos}W\\ \Rightarrow \mathrm{Cos}W=\frac{313-400}{312}\\ \Rightarrow \mathrm{Cos}W=\frac{-87}{312}\\ \Rightarrow \mathrm{Cos}W=-0.2788\\ \Rightarrow W={\cos }^{-1}(-0.2788)\\ \Rightarrow W={106.18}^{\circ }\\ \Rightarrow W={106}^{\circ }\end{array}$

As we know

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

So

  $\begin{array}{l}W+X+Y={180}^{\circ }\\ \Rightarrow Y={180}^{\circ }-(W+X)\\ \Rightarrow Y={180}^{\circ }-({106}^{\circ }+{39}^{\circ })\\ \Rightarrow Y={180}^{\circ }-{145}^{\circ }\\ \Rightarrow Y{=}^{}{35}^{\circ }\end{array}$

In triangle WXY

Angle $\angle W={106}^{\circ }$

Angle $\angle X={39}^{\circ }$

Angle $\angle Y={35}^{\circ }$