Chapter 8.6, Problem 11CYU

To determine

To calculate:

The each angle of triangle $\Delta DEF$

Answer to Problem 11CYU

The angles are $m\angle D={73}^{\circ },m\angle E={62}^{\circ },m\angle F={45}^{\circ }$ .

Explanation of Solution

Given information:

  $\begin{array}{l}DE=16\\ EF=21.6\\ FD=20\end{array}$

Concept used:

Law of Cosines:

  $c^2=a^2+b^2-2ab\cos C$

Calculation:

Since, the given three sides and no angles start solving this triangle $m\angle E$ using the Law of Cosines:

Law of Cosines:

  $c^2=a^2+b^2-2ab\cos C\mathrm{...}\left(1\right)$

Where,

  $\begin{array}{l}a=DE=16\\ b=EF=21.6\\ c=FD=20\\ C=E\end{array}$

Put the values in equation (1):

  ${\left(20\right)}^2={\left(16\right)}^2+{\left(21\cdot 6\right)}^2-2\left(16\right)\left(21\cdot 6\right)\cos E$

Simplify the equation:

  $400=256+466.56-691.2\cos E$

Use the inverse cosine ratio:

  $\begin{array}{l}3.8087\approx \cos E\\ 62\approx E\end{array}$

Similarly, use the Law of cosines for $m\angle D$ :

Where,

  $\begin{array}{l}a=DE=16\\ b=FD=20\\ c=EF=21.6\\ C=D\end{array}$

Put the values in equation (1):

  ${\left(21.6\right)}^2={\left(16\right)}^2+{\left(20\right)}^2-2\left(16\right)\left(20\right)\cos D$

Simplify the equation:

  $466.56=256+400-640\cos D$

Use the inverse cosine ratio:

  $\begin{array}{l}0.296\approx \cos D\\ 73\approx D\end{array}$

The sum of the angle of the triangle is $180$ .

  $m\angle F=180-\left(73+62\right)=45$

The angle of each triangle $m\angle D={73}^{\circ },m\angle E={62}^{\circ },m\angle F={45}^{\circ }$