Chapter 8.4, Problem 26AYU

Area of a Triangle Prove the two other forms of the formula given in Problem 27.

$K=\frac{b^2\sin A\sin C}{2\sin B}\text{and}K=\frac{c^2\sin A\sin B}{2\sin C}$

To determine

To prove: $K = \frac{b^2\text{sin}A\text{sin}C}{2\text{sin}B}$ and $K = \frac{c^2\text{sin}A\text{sin}B}{2\text{sin}C}$

Answer to Problem 26AYU

Solution:

Explanation of Solution

Given:

If two angles and the included side are given, the third angle is easy to find. The area $\text{K}$ of a triangle with side a and angles $\text{A}$, $\text{B}$, and $\text{C}$ is

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

Formula used:

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

Area of triangle, if sides b and c and the included angle $\text{A}$ are known is given by

$Area = \frac{1}{2} bc\text{sin}A$

Area of triangle, if sides a and c and the included angle $\text{B}$ are known is given by

$Area = \frac{1}{2} ac\text{sin}B$

Proof:

If the sides b, c and angle $\text{A}$ are known then,

$K = \frac{1}{2} bc\text{sin}A$   ––––– (1)

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

$c = \frac{b\text{sin}C}{\text{sin}B}$   ––––– (2)

Substituting (2) in (1) we get,

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

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

If the sides a, c and angle $\text{B}$ are known then,

$K = \frac{1}{2} ac\text{sin}B$   ––––– (1)

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

$a = \frac{c\text{sin}A}{\text{sin}C}$   ––––– (2)

Substituting (2) in (1) we get,

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

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