Chapter 14.3, Problem 24PPSE

Solution

To find: the remaining sides and angles of the given triangle.

  $A={50.6}^{\circ },B={39.4}^{\circ },b\approx 14$ .

Given:

  $a=17,c=22$ .

Concept used:

The sum of two acute angles of a right triangle is complementary.

Pythagorean Theorem: In any right triangle, the sum of squares of the lengths of two legs is equal to the

Square of the hypotenuse.

Calculation:

Let us draw the right triangle by using the given values as follows:

  

In the above triangle,

  $\begin{array}{c}\cos B=\frac{17}{22}\\ B={39.4005}^{\circ }\\ B\approx {39.4}^{\circ }\end{array}$

Now since the sum of two acute angles of a right triangle is complementary, therefore

  $\begin{array}{c}A+B={90}^{\circ }\\ A+{39.4}^{\circ }={90}^{\circ }\\ A={50.6}^{\circ }\end{array}$

Now by using the Pythagorean Theorem in the triangle ABC, it follows that

  $\begin{array}{c}{17}^2+b^2={22}^2\\ 289+b^2=484\\ b^2=195\\ b=\sqrt{195}\\ b=13.96\\ b\approx 14\end{array}$