Chapter 5.6, Problem 54E

Solution

To prove: The area of the triangle can be determined by the formula $\text{Area}=\frac{a^2\sin B\sin C}{2\sin A}$ .

Given information:

The formula for area of triangle is $\text{Area}=\frac{a^2\sin B\sin C}{2\sin A}$ .

Formula used:

Area of a right angled triangle is as follows:

  $\text{Area}=\frac{1}{2}\times \text{base}\text{×}\text{height}$

Law of Sins is as follows:

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

Proof:

Consider a triangle ABC such that $AD$ is perpendicular to $BC$ .

  

For the triangle ABC, $\text{Area}=\frac{1}{2}\times AD\text{×}BC$

From the triangle ABD,

  $\begin{array}{c}\sin B=\frac{\text{opposite}}{\text{hypotenuse}}\\ =\frac{AD}{c}\\ AD=c\sin B\end{array}$

Rewrite the Law of Sine to express $c$ in terms of $A$ , $a$ , $\sin A$

  $\begin{array}{c}\frac{a}{\sin A}=\frac{c}{\sin C}\\ c=\frac{a\sin C}{\sin A}\end{array}$

Substitute $\frac{a\sin C}{\sin A}$ for $c$ in $AD=c\sin B$

  $AD=\frac{a\sin C}{\sin A}\sin B$

Substitute $a$ for $BC$ and $\frac{a\sin C}{\sin A}\sin B$ for $AD$ in $\text{Area}=\frac{1}{2}\times AD\text{×}BC$

  $\begin{array}{c}\text{Area}=\frac{1}{2}\times AD\text{×}BC\\ =\frac{1}{2}\times \frac{a\sin C}{\sin A}\sin B\times a\\ =\frac{a^2\sin B\sin C}{2\sin A}\end{array}$

Hence, proved.