Chapter 13.4, Problem 13PPS

To determine

To find: Area of $\Delta ABC$ .

Answer to Problem 13PPS

Area of $\Delta ABC$ $=A=10.6$ square kilometers.

Explanation of Solution

Given information:

  $\begin{array}{l}C={45}^0\\ a=5km\\ b=6km\end{array}$

Concept:

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

Given

  $\begin{array}{l}C={45}^0\\ a=5km\\ b=6km\end{array}$

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

On substituting values

We get

  $\begin{array}{l}A=\frac{1}{2}\left(ab\sin \left(C\right)\right)\\ \Rightarrow A=\frac{1}{2}\left(\left(5\right)\left(6\right)\sin \left({45}^0\right)\right)\\ \Rightarrow A=\frac{5\times 6\times \left(0.7071\right)}{2}\\ \Rightarrow A=\frac{21.2132}{2}\\ \Rightarrow A=10.6\end{array}$

Therefore,

Area of $\Delta ABC$ $=A=10.6$ square kilometers.