Question

what are the values of m and θ in the diagram below? (0,1) $\frac{pi}{2}$ ($\frac{1}{2}$,m) (-1,0) π θ 0 (1,0) $\frac{3pi}{2}$ (0,-1) $m = \frac{sqrt{3}}{2}$, $\theta=\frac{pi}{3}$

Explanation:

Step1: Use the unit - circle equation

The equation of a unit - circle is $x^{2}+y^{2}=1$. Given $x = \frac{1}{2}$, we substitute it into the equation: $(\frac{1}{2})^{2}+m^{2}=1$.

Step2: Solve for $m$

$\frac{1}{4}+m^{2}=1$, then $m^{2}=1 - \frac{1}{4}=\frac{3}{4}$, so $m=\pm\frac{\sqrt{3}}{2}$. Since the point $(\frac{1}{2},m)$ is in the first - quadrant, $m=\frac{\sqrt{3}}{2}$.

Step3: Find the angle $\theta$

We know that $\cos\theta=x=\frac{1}{2}$ and $\sin\theta = m=\frac{\sqrt{3}}{2}$. In the range $[0,2\pi]$, when $\cos\theta=\frac{1}{2}$ and $\sin\theta=\frac{\sqrt{3}}{2}$, $\theta=\frac{\pi}{3}$.

Answer:

$m = \frac{\sqrt{3}}{2},\theta=\frac{\pi}{3}$