Chapter 12.5, Problem 15PPS

To determine

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

Answer to Problem 15PPS

In triangle FGH

Angle $\angle F={65}^{\circ }$

Angle $\angle G={94}^{\circ }$

Angle $\angle H={21}^{\circ }$

Explanation of Solution

Given information:

The length three sides of triangle FGH.

  GH= f = 10

  FH= g = 11

  FG= h = 4

Calculation:

To find the angle F, G and H we use the cosine formula for any triangle:

  $\begin{array}{l}{g}^2=f^2+h^2-2fh\cos G\\ \Rightarrow {11}^2={10}^2+4^2-2\times 10\times 4\times \cos G\\ \Rightarrow 121=100+16-80\mathrm{Cos}G\\ \Rightarrow 121=116-80\mathrm{Cos}G\\ \Rightarrow \mathrm{Cos}G=\frac{116-121}{80}\\ \Rightarrow \mathrm{Cos}G=\frac{-5}{80}\\ \Rightarrow \mathrm{Cos}G=-0.0625\\ \Rightarrow G={\cos }^{-1}(-0.0625)\\ \Rightarrow G=93.58\\ \Rightarrow G={94}^{\circ }\end{array}$

Also as we know that-

  $\begin{array}{l}{f}^2=g^2+h^2-2gh\cos F\\ \Rightarrow {10}^2={11}^2+4^2-2\times 11\times 4\times \cos F\\ \Rightarrow 100=121+16-88\mathrm{Cos}F\\ \Rightarrow 100=137-88\mathrm{Cos}F\\ \Rightarrow \mathrm{Cos}F=\frac{137-100}{88}\\ \Rightarrow \mathrm{Cos}F=\frac{37}{80}\\ \Rightarrow \mathrm{Cos}F=0.4625\\ \Rightarrow F={\cos }^{-1}(0.4204)\\ \Rightarrow F={65.13}^{\circ }\\ \Rightarrow F={65}^{\circ }\end{array}$

As we know

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

So

  $\begin{array}{l}F+G+H={180}^{\circ }\\ \Rightarrow H={180}^{\circ }-(F+G)\\ \Rightarrow H={180}^{\circ }-({65}^{\circ }+{94}^{\circ })\\ \Rightarrow H={180}^{\circ }-{159}^{\circ }\\ \Rightarrow H{=}^{}{21}^{\circ }\\ \end{array}$

In triangle FGH

Angle $\angle F={65}^{\circ }$

Angle $\angle G={94}^{\circ }$

Angle $\angle H={21}^{\circ }$