Chapter 8.4, Problem 62HP

To determine

To calculate:

The $△ABC$ , round to the nearest whole number.

Answer to Problem 62HP

The values of $△ABC$ are $m\angle A=53,m\angle B=90,m\angle C=37,AB=12,BC=16,AC=20$ .

Explanation of Solution

Given information:

  

Calculation:

With the help of the given figure it is easy to measure angle.

The sum of the measures of the angles of a triangle is $180$ .

Triangle, angle sum theorem shows that

  $\begin{array}{l}\left(7x+1\right)+\left(4x+1\right)+\left(3x-2\right)=180\\ 14x-2=180\\ 14x=182\\ x=13\end{array}$

Now find the measure of each angle:

  $\begin{array}{l}m\angle A=4x+1\\ =4\left(13\right)+1\\ =53\\ m\angle B=7x-1\\ =7\left(13\right)-1\\ =90\\ m\angle C=3x-2\\ =3\left(13\right)-2\\ =37\end{array}$

Thus, $△ABC$ is a right triangle with the right angle $B$ .

In a right triangle, the sum of the squares of the lengths of the lengths of the legs is equal to the square of the length of the hypotenuse.

Pythagoras Theorem shows that,

  $\begin{array}{l}{\left(3y-1\right)}^2={\left(y+5\right)}^2+{\left(2y+2\right)}^2\\ 9y^2-6y+1=y^2+10y+25+4y^2+8y+4\\ 9y^2-6y+1=5y^2+18y+29\\ 4y^2-4y-28=0\\ y^2-y-7=0\end{array}$

Now, use quadratic formula to solve the resulting quadratic equation.

  $\begin{array}{l}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(-7\right)}}{2\left(1\right)}\\ =\frac{6\pm 8}{2}\\ =-1or7\end{array}$

The negative solution $y=-1$ is extraneous because side length must be positive.

Find the measure of each side,

  $\begin{array}{l}AB=y+5\\ =7+5\\ =12\\ BC=2y+2\\ =2\left(7\right)+2\\ =16\\ AC=3y-1\\ =3\left(7\right)-1\\ =20\end{array}$

Hence, $m\angle A=53,m\angle B=90,m\angle C=37,AB=12,BC=16,AC=20$ .