Question

with a graphing calculator, find the coordinates of the point of intersection of y = \sqrt{3} \cdot x and y = \sqrt{1 - x^{2}}. these coordinates are the cosine and sine of what angle between 0° and 90°?
(type a whole number.)

Explanation:

Step1: Set the two equations equal

Set $\sqrt{3}x=\sqrt{1 - x^{2}}$. Square both sides to get $3x^{2}=1 - x^{2}$ (note: we need to check for extraneous solutions later since squaring can introduce them).

Step2: Solve for $x$

Combine like - terms: $3x^{2}+x^{2}=1$, so $4x^{2}=1$. Then $x^{2}=\frac{1}{4}$, and $x=\pm\frac{1}{2}$. Since we are looking for the intersection in the first - quadrant (because we want an angle between $0^{\circ}$ and $90^{\circ}$), we take $x = \frac{1}{2}$.

Step3: Find the $y$ - value

Substitute $x=\frac{1}{2}$ into $y = \sqrt{3}x$, so $y=\sqrt{3}\times\frac{1}{2}=\frac{\sqrt{3}}{2}$.

Step4: Find the angle

We know that if $x=\cos\theta$ and $y = \sin\theta$, and $x=\frac{1}{2}$, $y=\frac{\sqrt{3}}{2}$, then $\theta = 60^{\circ}$ (because $\cos60^{\circ}=\frac{1}{2}$ and $\sin60^{\circ}=\frac{\sqrt{3}}{2}$).

Answer:

$60$