In this activity, you might have noticed the following:

• The radius of the circle is  \(r=5\)​​.​​​​​
• Moving point \(P\) changes the angle \(\theta\).
• For \(P\left(x,y\right)\) on the terminal arm of \(\theta\), 
  • the sine ratio is \(\sin\left(\theta\right)=\dfrac{y}{r}\);
  • the cosine ratio is \(\cos\left(\theta\right)=\dfrac{x}{r}\); and
  • the tangent ratio is \(\tan\left(\theta\right)=\dfrac{y}{x}\).

If you like the mnemonic SOH CAH TOA, you might use the more sophisticated SYR CXR TYX, pronounced “sir kixer ticks.”

Solution — Part A

For point \(A\left(-2,7\right)\),  ​\(x=-2\), \(y=7\), and

\(r=\sqrt{x^2+y^2}\class{timed in7}{=\sqrt{\left(-2\right)^2+7^2}=\sqrt{53}}\).

Therefore,

• \(\sin \left(\theta \right) = \dfrac{y}{r}= \dfrac{7}{\sqrt{53}} \)
• \(\cos \left(\theta \right) = \dfrac{x}{r}= -\dfrac{2}{\sqrt{53}}\)
• \(\tan \left(\theta \right) = \dfrac{y}{x}= -\dfrac{7}{2} \)

Solution — Part B

For point \(B\left(-6,-2\right)\), \(x=-6\), \(y=-2\), and

\(r=\sqrt{\left(-6\right)^2+\left(-2\right)^2}=\sqrt{40}\class{timed in3}{=\sqrt{4\times 10}=2\sqrt{10}}\).

Therefore,

Solution — Part C

For point \(C\left(0,-3\right)\), \(x=0\), \(y=-3\), and

\(r=\sqrt{0^2+\left(-3\right)^2}=3\).

Since \(C\) is on the negative \(y\)-axis, we know \(\theta=270^\circ\). 

Therefore,

\(\sin \left(270^\circ \right) =\dfrac{y}{r} = \dfrac{-3}{3}=-1 \)

\( \cos \left(270^\circ \right) =\dfrac{x}{r} = \dfrac{0}{3}=0 \)

\(\tan \left(270^\circ \right) =\dfrac{y}{x} = \dfrac{-3}{0} \textrm{ is undefined} \)