Chapter 12.4, Problem 1GP

To determine

To find: Area of $\Delta ABC$ .

Answer to Problem 1GP

Area of $\Delta ABC$ $=A=102.0$ square meters.

Explanation of Solution

Given information:

  $\begin{array}{l}A={31}^0\\ b=18\\ c=22\end{array}$

Concept:

Area of Triangle $=A=\frac{1}{2}\left(bc\sin \left(A\right)\right)$

Given

  $\begin{array}{l}A={31}^0\\ b=18\\ c=22\end{array}$

Area of Triangle $=A=\frac{1}{2}\left(bc\sin \left(A\right)\right)$

On substituting values

We get

  $\begin{array}{l}A=\frac{1}{2}\left(bc\sin \left(A\right)\right)\\ \Rightarrow A=\frac{1}{2}\left(\left(18\right)\left(22\right)\sin \left({31}^0\right)\right)\\ \Rightarrow A=\frac{18\times 22\times \left(0.5150\right)}{2}\\ \Rightarrow A=\frac{203.955}{2}\\ \Rightarrow A=102.0\end{array}$

Therefore,

Area of $\Delta ABC$ $=A=102.0$ square meters.