Chapter 8.6, Problem 33PPS

To determine

To calculate:

The triangle and angle measures degree and side measures to the nearest tenth.

Answer to Problem 33PPS

The measurements of triangle are $m\angle J\approx {65}^{\circ },m\angle K\approx {66}^{\circ },m\angle L\approx {49}^{\circ }$ .

Explanation of Solution

Given information:

  

Concept used:

Law of cosines:

  $c^2=a^2+b^2-2ab\cos C$

Calculation:

  

With the help of the above diagram it is easy to get all three sides and no angle. So, with the help of the Law of cosines get the missing angle.

Use the law of cosines:

Law of Cosines:

  $c^2=a^2+b^2-2ab\cos C$

So, first substitute

  ${30.0}^2={\left(24.6\right)}^2+{\left(29.7\right)}^2-2\left(24.6\right)\left(29.7\right)\cos K$

Now, simplify

  $900=605.16+882.09-1461.24\cos K$

Now, solve for $K$ ,

  $0.40188=\cos K$

Now use the inverse cosine ratio to get the value of $K$ ,

  $K\approx 66$

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

So, substitute

  ${29.7}^2={\left(30\right)}^2+{\left(24.6\right)}^2-2\left(30\right)\left(24.6\right)\cos J$

Now, simplify

  $882.09=900+605.16-147\cos J$

Now, solve for $J$ ,

  $0.42213=\cos J$

Now use the inverse cosine ratio to get the value of $J$ ,

  $J\approx 65$

The sum of the angles of a triangle is $180$ . So, $m\angle L=180-\left(66+65\right)=49$

Therefore, the measurements of triangle are $m\angle J\approx {65}^{\circ },m\angle K\approx {66}^{\circ },m\angle L\approx {49}^{\circ }$