Chapter 13.7, Problem 37WE

To determine

To Find: The expression for the height $h$ .

Answer to Problem 37WE

The equation for the height $h=\frac{d\sin \alpha \sin \beta }{\sin \left(\alpha +\beta \right)}$ is derived.

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

The expression for the height is $h=\frac{d\sin \alpha \sin \beta }{\sin \left(\alpha +\beta \right)}$ .

Calculation:

Consider the triangle ABC,

  $\begin{array}{l}\tan \left(180-\alpha \right)=\frac{h}{x}\\ -\tan \alpha =\frac{h}{x}\\ -\tan \alpha \left(\frac{x}{\tan \alpha }\right)=\frac{h}{x}\left(\frac{x}{\tan \alpha }\right)\\ x=-\frac{h}{\tan \alpha }\end{array}$

Consider the triangle

  $ABD$ as,

  $\begin{array}{l}\tan \beta =\frac{h}{x+d}\\ \tan \beta \left(\frac{x+d}{\tan \beta }\right)=\frac{h}{x+d}\left(\frac{x+d}{\tan \beta }\right)\\ x+d=\frac{h}{\tan \beta }\end{array}$

Then, substitute $x=-\frac{h}{\tan \alpha }$ in the above equation as,

  $\begin{array}{l}-\frac{h}{\tan \alpha }+d=\frac{h}{\tan \beta }\\ d=h\left(\frac{1}{\tan \alpha }+\frac{1}{\tan \beta }\right)\\ d=h\left(\frac{\cos \alpha }{\sin \alpha }+\frac{\cos \beta }{\sin \beta }\right)\\ d=h\left(\frac{\cos \alpha \sin \beta +\sin \alpha \cos \beta }{\sin \alpha \sin \beta }\right)\end{array}$

Solve further as, by the trigonometric identity as,

  $\begin{array}{l}d=h\left(\frac{\sin \left(\alpha +\beta \right)}{\sin \alpha \sin \beta }\right)\\ h=\frac{d\sin \alpha \sin \beta }{\sin \left(\alpha +\beta \right)}\end{array}$

Thus, the equation for the height

  $h=\frac{d\sin \alpha \sin \beta }{\sin \left(\alpha +\beta \right)}$ is derived.