Chapter 8.4, Problem 25AYU

Area of an ASA Triangle If two angles and the included side are given, the third angle is easy to find. Use the Law of Sines to show that the area $K$ of a triangle with side $a$ and angles $A,B,$ and $C$ is

$K=\frac{a^2\sin B\sin C}{2\sin A}$

To determine

To show: That the area $\text{K}$ of a triangle with side a and angles $\text{A}$, $\text{B}$ and $\text{C}$ is

$K = \frac{a^2\text{sin}B\text{sin}C}{2\text{sin}A}$

Answer to Problem 25AYU

Solution:

The area $\text{K}$ of a triangle with side a and angles $\text{A}$, $\text{B}$ and $\text{C}$ is

$K = \frac{a^2\text{sin}B\text{sin}C}{2\text{sin}A}$

Explanation of Solution

Given:

If two angles and the included side are given, the third angle is easy to find. Use the Law of Sines.

Formula used:

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

$Area = \frac{1}{2} ab\text{sin}C$

Proof:

$K = \frac{1}{2} ab\text{sin}C$   -----(1)

$\frac{\text{sin}A}{a} = \frac{\text{sin}B}{b}$

$b = \frac{a\text{sin}B}{\text{sin}A}$   -----(2)

Substituting (2) in (1) we get,

$K = \frac{1}{2} a \left(\frac{a\text{sin}B}{\text{sin}A}\right)\text{sin}C$

$K = \frac{1}{2} a^2\text{sin}B\frac{\text{sin}C}{\text{sin}A}$

$K = \frac{a^2\text{sin}B\text{sin}C}{2\text{sin}A}$