Chapter 6, Problem 16RE

To determine

The area of the triangle having the indicated angleand sides.

Answer to Problem 16RE

Thearea of triangle is $\text{44}\text{.08}$

Explanation of Solution

Given info:

  $A={11}^0,b=22,c=21$

Formula used:

  

Law of sines

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Area of triangle

  $A=\frac{1}{2}bc\sin A=\frac{1}{2}ac\sin B=\frac{1}{2}ab\sin C$

Calculation:

Given that $A={11}^0,b=22,c=21$ , We have

  $\begin{array}{l}A=\frac{1}{2}bc\sin A=\frac{1}{2}ac\sin B=\frac{1}{2}ab\sin C\\ A=\frac{1}{2}bc\sin A=\frac{1}{2}\times 22\times 21\times \sin {11}^0=\text{44}\text{.08}\end{array}$

Conclusion:

Thus, the area of triangle is $\text{44}\text{.08}$