Chapter 13.4, Problem 18PPS

To determine

To find: Area of $\Delta ABC$ .

Answer to Problem 18PPS

Area of $\Delta ABC$ $=A=66.9$ square inch.

Explanation of Solution

Given information:

  $\begin{array}{l}A={138}^0\\ b=10in\\ c=20in\end{array}$

Concept:

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

Given

  $\begin{array}{l}A={138}^0\\ b=10in\\ c=20in\end{array}$

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

On substituting values

We get

  $\begin{array}{l}A=\frac{1}{2}\left(bc\sin \left(A\right)\right)\\ \Rightarrow A=\frac{1}{2}\left(\left(10\right)\left(20\right)\sin \left({138}^0\right)\right)\\ \Rightarrow A=\frac{10\times 20\times \left(0.67\right)}{2}\\ \Rightarrow A=\frac{133.826}{2}\\ \Rightarrow A=66.9\end{array}$

Therefore,

Area of $\Delta ABC$ $=A=66.9$ square inch.