Chapter 8.6, Problem 40PPS

To determine

To calculate:

The $△JKL$ for given measures.

Answer to Problem 40PPS

The measurements of triangle are $m\angle L\approx {31}^{\circ },m\angle K\approx {90}^{\circ },m\angle E\approx {59}^{\circ }$

Explanation of Solution

Given information:

Measurements of $△JKL$ :

  $\begin{array}{l}JK=33\\ KL=56\\ LJ=65\end{array}$

Concept used:

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Where,

  $a,b=$ sides

  $\cos C=$ angle

Calculation:

The given measurements of $△JKL$ are $JK=33,KL=56,LJ=65$ .

So, it is easy to get the three sides of triangle and non-included angle. So, with the help of the Law of cosines get the missing angle.

First consider $K$ to be the angle.

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Where,

  $\begin{array}{l}a=56\\ b=33\\ c=65\\ \cos C=K\end{array}$

First substitute,

  ${\left(65\right)}^2={\left(56\right)}^2+{\left(33\right)}^2-2\left(56\right)\left(33\right)\cos \left(K\right)$

Now simplify,

  $4225=3136+1089-3696\cos K$

Now solve for the $K$

  $0=\cos K$

Now, get the value of $K$ by using the inverse cosine ratio

  $K\approx 90$

Similarly, use the law of cosine to get the measure of angle $J$

First substitute,

  ${\left(56\right)}^2={\left(33\right)}^2+{\left(65\right)}^2-2\left(33\right)\left(65\right)\cos J$

Now simplify,

  $3136=1089+4225-4290\cos J$

Now solve for the $J$

  $0.5077\approx \cos J$

Now, get the value of $J$ by using the inverse cosine ratio

  $J\approx 59$

The sum of the angles of a triangle is $180$ . So,

  $m\angle L=180-\left(90+59\right)=31$

Hence, the measurements of triangle are $m\angle L\approx {31}^{\circ },m\angle K\approx {90}^{\circ },m\angle E\approx {59}^{\circ }$