Chapter 13.5, Problem 1CYP

To determine

To find: To round side lengths to the nearest tenth and angles to the nearest degree.

Answer to Problem 1CYP

In triangle FGH

So Angle $\angle F={62}^{\circ }$

Angle $\angle H={36}^{\circ }$

And side FH= g =6.7

Explanation of Solution

Given information:

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

GH = f = 6

FG= h= 4

∠FGH=∠G= 92

Calculation:

To find the length of side FH= g , we use the cosine formula for any triangle: $\begin{array}{l}{g}^2=f^2+h^2-2fh\cos G\\ \Rightarrow g^2=6^2+4^2-2\times 6\times 4\times \cos {82}^{\circ }\\ \Rightarrow g^2=36+16-48\mathrm{Cos}{82}^{\circ }\\ \Rightarrow g^2=52-6.6768\\ \Rightarrow g^2=52-6.6768\\ \Rightarrow g^2=45.32\\ \Rightarrow g=\sqrt{45.32}\\ \Rightarrow g=6.7\end{array}$

We also know that-

  $\begin{array}{l}\frac{\mathrm{Sin}F}{f}=\frac{\mathrm{Sin}G}{g}\\ \Rightarrow \mathrm{Sin}F=\frac{f\times \mathrm{Sin}82}{6.7}\\ \Rightarrow \mathrm{Sin}F=\frac{6\times \mathrm{Sin}82}{6.7}\\ \Rightarrow \mathrm{Sin}F=\frac{6\times 0.99}{6.7}\\ \Rightarrow \mathrm{Sin}F=0.8865\\ \Rightarrow F={\sin }^{-1}(0.8865)\\ \Rightarrow F={62.44}^{\circ }\\ \Rightarrow F={62}^{\circ }\end{array}$

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

So

  $\begin{array}{l}G+F+H={180}^{\circ }\\ \Rightarrow H={180}^{\circ }-(G+F)\\ \Rightarrow H={180}^{\circ }-({82}^{\circ }+{62}^{\circ })\\ \Rightarrow H={180}^{\circ }-{144}^{\circ }\\ \Rightarrow H={36}^{\circ }\end{array}$

In triangle FGH

So Angle $\angle F={62}^{\circ }$

Angle $\angle H={36}^{\circ }$

And side FH= g =6.7