Chapter 12.2, Problem 32WE

To determine

Use similar triangles to show that $\frac{y}{r}=\frac{y\text{'}}{s},\frac{x}{r}=\frac{x\text{'}}{s},\frac{y}{x}=\frac{y\text{'}}{x\text{'}}$

Explanation of Solution

Given:

  $P(x,y)$ and $Q(x\text{'},y\text{'})$ are two points on the terminal side of an acute angle $\theta$ in standard position . Let OP = r and OQ = s .

Calculation:

Consider the triangle:

  

OP = r and OQ = s .

  $\begin{array}{l}OC=x\text{'},QC=y\text{'}\\ OB=x,PB=y\end{array}$

  $\Delta OPB\simeq \Delta OQC$

So,

  $\begin{array}{l}\frac{OP}{OQ}=\frac{PB}{QC}=\frac{OB}{OC}\\ \text{⇒}\frac{r}{s}=\frac{y}{y\text{'}}=\frac{x}{x\text{'}}\mathrm{................}[OP=r,OQ=s;PB=y,QC=y\text{'};OB=x,OC=x\text{'}]\end{array}$

Since

  $\begin{array}{l}\frac{r}{s}=\frac{y}{y\text{'}}\\ \text{⇒}ry\text{'}=ys\mathrm{...............}[\text{multiply each side by sy']}\\ \text{⇒}\frac{y\text{'}}{s}=\frac{y}{r}\mathrm{...............}[\text{divide each side by rs]}\\ \text{⇒}\frac{y}{r}=\frac{y\text{'}}{s}\mathrm{....................}(1)\end{array}$

Since

  $\begin{array}{l}\frac{y}{y\text{'}}=\frac{x}{x\text{'}}\\ \text{⇒}yx\text{'}=xy\text{'}\mathrm{.......................}[\text{multiply by x'y']}\\ \text{⇒}\frac{y}{x}=\frac{y\text{'}}{x\text{'}}\mathrm{.............}(2)\mathrm{..............}[\text{divide each side by xx']}\end{array}$

Since

  $\begin{array}{l}\frac{r}{s}=\frac{x}{x\text{'}}\\ \text{⇒}rx\text{'}=xs\mathrm{.................}[\text{multiply each side by sx']}\\ \text{⇒}\frac{x\text{'}}{s}=\frac{x}{r}\mathrm{.................}[\text{divide each side by rs]}\\ \text{⇒}\frac{x}{r}=\frac{x\text{'}}{s}\mathrm{...................}(3)\end{array}$

Hence , from equation (1),(2) and (3):

  $\frac{y}{r}=\frac{y\text{'}}{s},\frac{x}{r}=\frac{x\text{'}}{s},\frac{y}{x}=\frac{y\text{'}}{x\text{'}}$

Hence proved.