Chapter 8.4, Problem 49AYU

To determine

To Prove:-The area for the altitude $h$ from the vertex to the opposite side $a$ of the triangle.

Explanation of Solution

Given:

The given formula for the altitude is,

  $h=\frac{a\sin B\sin C}{\sin A}$

Calculation:

Consider the area of the triangle is,

  $K=\frac{1}{2}ah$

Consider the area of the triangle with angle is,

  $K=\frac{1}{2}ab\sin C$

Then,

  $\begin{array}{l}\frac{1}{2}ah=\frac{1}{2}ab\sin C\\ h=b\sin C\end{array}$

Consider the law of sine is,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ b=\frac{a\sin B}{\sin A}\end{array}$

Then, $h=b\sin C$ is,

  $\begin{array}{c}h=\left(\frac{a\sin B}{\sin A}\right)\sin C\\ =\frac{a\sin B\sin C}{\sin A}\end{array}$

Hence, proved.