Question

sketch an angle θ in standard position such that θ has the least possible positive measure, and the point ((2, sqrt{3})) is on the terminal side of θ. then find the values of the six trigonometric functions for the angle. rationalize denominators if applicable. do not use a calculator. choose the correct graph below. ○ a. graph, ○ b. graph, ○ c. graph, ○ d. graph

Explanation:

Part 1: Choosing the Correct Graph

The point \((2, \sqrt{3})\) is in the first quadrant (since both \(x = 2\) and \(y=\sqrt{3}\) are positive). So the terminal side of the angle \(\theta\) should lie in the first quadrant.

• Option A: The point seems to be in the fourth quadrant (negative \(y\)-coordinate? Or check the grid—probably not first quadrant).
• Option B: The point is in the second quadrant (negative \(x\), positive \(y\))—incorrect.
• Option C: The point is in the fourth quadrant (positive \(x\), negative \(y\))—incorrect.
• Option D: The point is in the first quadrant (positive \(x\), positive \(y\))—matches \((2, \sqrt{3})\).

So the correct graph is D.

Part 2: Finding the Six Trigonometric Functions

To find the trigonometric functions, we first find \(r\) (the distance from the origin to the point \((x, y)\)) using the formula \(r=\sqrt{x^2 + y^2}\).

Step 1: Calculate \(r\)

Given \(x = 2\) and \(y=\sqrt{3}\):
\[
r=\sqrt{(2)^2+(\sqrt{3})^2}=\sqrt{4 + 3}=\sqrt{7}
\]

Step 2: Sine Function (\(\sin\theta\))

The formula for sine is \(\sin\theta=\frac{y}{r}\):
\[
\sin\theta=\frac{\sqrt{3}}{\sqrt{7}}=\frac{\sqrt{21}}{7} \quad (\text{after rationalizing the denominator: } \frac{\sqrt{3}\cdot\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}=\frac{\sqrt{21}}{7})
\]

Step 3: Cosine Function (\(\cos\theta\))

The formula for cosine is \(\cos\theta=\frac{x}{r}\):
\[
\cos\theta=\frac{2}{\sqrt{7}}=\frac{2\sqrt{7}}{7} \quad (\text{rationalizing: } \frac{2\cdot\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}=\frac{2\sqrt{7}}{7})
\]

Step 4: Tangent Function (\(\tan\theta\))

The formula for tangent is \(\tan\theta=\frac{y}{x}\) (since \(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{y/r}{x/r}=\frac{y}{x}\)):
\[
\tan\theta=\frac{\sqrt{3}}{2}
\]

Step 5: Cosecant Function (\(\csc\theta\))

The formula for cosecant is \(\csc\theta=\frac{r}{y}\) (reciprocal of sine):
\[
\csc\theta=\frac{\sqrt{7}}{\sqrt{3}}=\frac{\sqrt{21}}{3} \quad (\text{rationalizing: } \frac{\sqrt{7}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{21}}{3})
\]

Step 6: Secant Function (\(\sec\theta\))

The formula for secant is \(\sec\theta=\frac{r}{x}\) (reciprocal of cosine):
\[
\sec\theta=\frac{\sqrt{7}}{2}
\]

Step 7: Cotangent Function (\(\cot\theta\))

The formula for cotangent is \(\cot\theta=\frac{x}{y}\) (reciprocal of tangent):
\[
\cot\theta=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3} \quad (\text{rationalizing: } \frac{2\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{2\sqrt{3}}{3})
\]

Final Answers

Correct Graph:

\(\boldsymbol{\text{D}}\)

Trigonometric Functions:

\[

$$\begin{align*} \sin\theta &= \frac{\sqrt{21}}{7}, \\ \cos\theta &= \frac{2\sqrt{7}}{7}, \\ \tan\theta &= \frac{\sqrt{3}}{2}, \\ \csc\theta &= \frac{\sqrt{21}}{3}, \\ \sec\theta &= \frac{\sqrt{7}}{2}, \\ \cot\theta &= \frac{2\sqrt{3}}{3} \end{align*}$$

\]

Answer:

Part 1: Choosing the Correct Graph

The point \((2, \sqrt{3})\) is in the first quadrant (since both \(x = 2\) and \(y=\sqrt{3}\) are positive). So the terminal side of the angle \(\theta\) should lie in the first quadrant.

• Option A: The point seems to be in the fourth quadrant (negative \(y\)-coordinate? Or check the grid—probably not first quadrant).
• Option B: The point is in the second quadrant (negative \(x\), positive \(y\))—incorrect.
• Option C: The point is in the fourth quadrant (positive \(x\), negative \(y\))—incorrect.
• Option D: The point is in the first quadrant (positive \(x\), positive \(y\))—matches \((2, \sqrt{3})\).

So the correct graph is D.

Part 2: Finding the Six Trigonometric Functions

To find the trigonometric functions, we first find \(r\) (the distance from the origin to the point \((x, y)\)) using the formula \(r=\sqrt{x^2 + y^2}\).

Step 1: Calculate \(r\)

Given \(x = 2\) and \(y=\sqrt{3}\):
\[
r=\sqrt{(2)^2+(\sqrt{3})^2}=\sqrt{4 + 3}=\sqrt{7}
\]

Step 2: Sine Function (\(\sin\theta\))

The formula for sine is \(\sin\theta=\frac{y}{r}\):
\[
\sin\theta=\frac{\sqrt{3}}{\sqrt{7}}=\frac{\sqrt{21}}{7} \quad (\text{after rationalizing the denominator: } \frac{\sqrt{3}\cdot\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}=\frac{\sqrt{21}}{7})
\]

Step 3: Cosine Function (\(\cos\theta\))

The formula for cosine is \(\cos\theta=\frac{x}{r}\):
\[
\cos\theta=\frac{2}{\sqrt{7}}=\frac{2\sqrt{7}}{7} \quad (\text{rationalizing: } \frac{2\cdot\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}=\frac{2\sqrt{7}}{7})
\]

Step 4: Tangent Function (\(\tan\theta\))

The formula for tangent is \(\tan\theta=\frac{y}{x}\) (since \(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{y/r}{x/r}=\frac{y}{x}\)):
\[
\tan\theta=\frac{\sqrt{3}}{2}
\]

Step 5: Cosecant Function (\(\csc\theta\))

The formula for cosecant is \(\csc\theta=\frac{r}{y}\) (reciprocal of sine):
\[
\csc\theta=\frac{\sqrt{7}}{\sqrt{3}}=\frac{\sqrt{21}}{3} \quad (\text{rationalizing: } \frac{\sqrt{7}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{21}}{3})
\]

Step 6: Secant Function (\(\sec\theta\))

The formula for secant is \(\sec\theta=\frac{r}{x}\) (reciprocal of cosine):
\[
\sec\theta=\frac{\sqrt{7}}{2}
\]

Step 7: Cotangent Function (\(\cot\theta\))

The formula for cotangent is \(\cot\theta=\frac{x}{y}\) (reciprocal of tangent):
\[
\cot\theta=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3} \quad (\text{rationalizing: } \frac{2\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{2\sqrt{3}}{3})
\]

Final Answers

Correct Graph:

\(\boldsymbol{\text{D}}\)

Trigonometric Functions:

\[

$$\begin{align*} \sin\theta &= \frac{\sqrt{21}}{7}, \\ \cos\theta &= \frac{2\sqrt{7}}{7}, \\ \tan\theta &= \frac{\sqrt{3}}{2}, \\ \csc\theta &= \frac{\sqrt{21}}{3}, \\ \sec\theta &= \frac{\sqrt{7}}{2}, \\ \cot\theta &= \frac{2\sqrt{3}}{3} \end{align*}$$

\]