Chapter 12.7, Problem 22WE

To determine

Prove that $\frac{a}{b}=\frac{x}{y}$

Explanation of Solution

Formula Used:

  $\sin (180°-\theta )=\sin \theta$

Calculation:

Consider the triangle :

  

In $\Delta ANC\text{ and }\Delta BNC$ , apply law of sines:

  $\begin{array}{l}\frac{\sin \theta }{x}=\frac{\sin \alpha }{a};\frac{\sin \theta }{y}=\frac{\sin \beta }{b}\\ \Rightarrow \frac{\sin \theta }{x}=\frac{\sin \alpha }{a};\frac{\sin \theta }{y}=\frac{\sin (180°-\alpha )}{b}\mathrm{.........}[\beta =180°-\alpha ]\\ \Rightarrow \frac{\sin \theta }{x}=\frac{\sin \alpha }{a};\frac{\sin \theta }{y}=\frac{\sin \alpha }{b}\mathrm{.....................}[\sin (180°-\theta )=\sin \theta ]\\ \Rightarrow \frac{\sin \theta }{\sin \alpha }=\frac{x}{a};\frac{\sin \theta }{\sin \alpha }=\frac{y}{b}\\ \Rightarrow \frac{x}{a}=\frac{y}{b}=\frac{\sin \theta }{\sin \alpha }\\ \Rightarrow \frac{x}{a}=\frac{y}{b}\\ \Rightarrow \frac{x}{y}=\frac{a}{b}\mathrm{.............................................}[\text{multiply each side by }\frac{a}{y}]\end{array}$

Hence proved.