Chapter C, Problem 46E

(a)

To determine

To show:

  $A=\frac{1}{2}ab\sin \theta$

Explanation of Solution

Given information:

Length of two sides: $a,b$

Angle: $\theta$

Formula Used:

The area of triangle is given by the following expression:

  $A=\frac{1}{2}bh$

Where,

Base: $b$

Height: $h$

Proof:

First sketch the triangle with two side length as $a,b$ and includes an angle of $\theta$ . Then height $h$ as perpendicular to the side $b$ :

  

From figure:

  $\begin{array}{l}\sin \theta =\frac{h}{a}\\ h=a\sin \theta \end{array}$

So, the area of triangle is given by the following expression:

  $\begin{array}{l}A=\frac{1}{2}bh\\ A=\frac{1}{2}b\left(a\sin \theta \right)\\ A=\frac{1}{2}ab\sin \theta \end{array}$

Hence, proved.

(b)

To determine

To calculate:

The area of $\Delta ABC$

Answer to Problem 46E

The area of $\Delta ABC$ is $14.34457c{m}^2$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\left|AB\right|=10cm\\ \left|BC\right|=3cm\\ \angle ABC=107°\end{array}$

Formula Used:

The area of triangle is given by the following expression:

  $A=\frac{1}{2}bh$

Where,

Base: $b$

Height: $h$

Calculation:

  $\Delta ABC$ is drawn as follow with the given information:

  

The area of triangle is given by the following expression:

  $A=\frac{1}{2}bh$

Put the given data:

  $\begin{array}{l}A=\frac{1}{2}bh\\ A=\frac{1}{2}\left|AB\right|\left|BC\right|\angle ABC\\ A=\frac{1}{2}\left(10\times 3\times 107\right)\\ A=14.34457c{m}^2\end{array}$