Chapter 6.1, Problem 2QR

Solution

To find: The value of $x$ and $y$ .

  $\begin{array}{l}x=7.5\\ y=-13\end{array}$

Given information:

Find the value of $x$ and $y$ .

  

Concept used:

The angle is formed by the clockwise rotation. So, the angle is negative.

Use the trigonometric function of angle to find the value of $x$ and $y$ .

Explanation:

The distance of $\left(x,y\right)$ from the origin is $15$ .

Let the angle formed in reference triangle be $\theta$ .

Since the measure of straight angle is $180°$ , it follows that:

  $\begin{array}{l}\theta +120°=180°\\ \Rightarrow \theta =60°\end{array}$

Since the angle is formed by the clockwise rotation, the angle is $-60°$ .

The perpendicular side in reference triangle is $y$ , the adjacent side is $x$ and the hypotenuse is $15$ .

Find $y$ .

  $\begin{array}{l}\sin \left(-60°\right)=\frac{\text{perpendicular}\text{side}}{\text{hypotenuse}}\\ \Rightarrow -\frac{\sqrt{3}}{2}=\frac{y}{15}\left[\sin \left(-60°\right)=-\frac{\sqrt{3}}{2}\right]\\ \Rightarrow y=-\frac{15\sqrt{3}}{2}\left[\text{Take}\sqrt{3}=1.732\right]\\ \Rightarrow y=-\frac{15\times 1.732}{2}=-12.99\\ \Rightarrow y=-13\end{array}$

Find $x$ .

  $\begin{array}{l}\cos \left(-60°\right)=\frac{\text{adjacent}\text{side}}{\text{hypotenuse}}\\ \Rightarrow \frac{1}{2}=\frac{x}{15}\left[\cos \left(-60°\right)=\frac{1}{2}\right]\\ \Rightarrow x=\frac{15}{2}=7.5\end{array}$