Upload your work... Draw the triangle in the correct quadrant with ALL appropriate labels. This includes finding the MISSING SIDE. If \( \sin \theta=1 / 3 \) and \( \theta \) is in Quad ॥

Certainly! Let's work through the problem step by step to draw the appropriate right triangle, label all parts, determine the missing side, and indicate the correct quadrant for angle \(\theta\). ### Given: - \(\sin \theta = \frac{1}{3}\) - \(\theta\) is in **Quadrant II**. ### Understanding the Quadrant: In **Quadrant II**: - **Sine** values are **positive**. - **Cosine** and **Tangent** values are **negative**. ### Step 1: Draw the Coordinate Plane and Identify Quadrant II ``` | | Quadrant II | | |------------------------- | | ``` ### Step 2: Draw the Right Triangle in Quadrant II 1. **Angle \(\theta\)** is positioned such that its vertex is at the origin, and its terminal side extends into Quadrant II. 2. Draw a perpendicular from the terminal side to the x-axis to form a right triangle. ``` | |\ | \ 1 | \ 3 | \ |____\ -x ``` - **Opposite Side (Vertical side)**: 1 (since \(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{3}\)) - **Hypotenuse**: 3 - **Adjacent Side (Horizontal side)**: To be determined (denoted as \(-x\) because in Quadrant II, the adjacent side is negative) ### Step 3: Determine the Missing Side Using the Pythagorean Theorem The Pythagorean Theorem states: \[ \text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2 \] Plugging in the known values: \[ 3^2 = 1^2 + (-x)^2 \\ 9 = 1 + x^2 \\ x^2 = 9 - 1 \\ x^2 = 8 \\ x = \sqrt{8} = 2\sqrt{2} \] Since the adjacent side is in Quadrant II, it's negative: \[ \text{Adjacent} = -2\sqrt{2} \] ### Step 4: Label All Parts of the Triangle ``` | |\ | \ 1 | \ 3 | \ |____\ -2√2 x-axis ``` - **Angle \(\theta\)**: Located at the origin. - **Opposite Side**: 1 (vertical side) - **Adjacent Side**: \(-2\sqrt{2}\) (horizontal side, negative in Quadrant II) - **Hypotenuse**: 3 ### Final Diagram Description - **Origin (0,0)**: Vertex of angle \(\theta\). - **Terminal Side of \(\theta\)**: Extends into Quadrant II. - **Opposite Side**: Length of 1, along the y-axis. - **Adjacent Side**: Length of \(2\sqrt{2}\), extending into the negative direction of the x-axis. - **Hypotenuse**: Connecting the origin to the point \((-2\sqrt{2}, 1)\), length 3. ### Summary - **\(\sin \theta = \frac{1}{3}\)** implies: - **Opposite Side**: 1 - **Hypotenuse**: 3 - **Using the Pythagorean Theorem**, the **Adjacent Side** is found to be \(-2\sqrt{2}\) (negative in Quadrant II). - All parts are labeled appropriately in the right triangle positioned in **Quadrant II**. If you need a visual diagram, here's a textual representation: ``` Quadrant II | |\ | \ 1 | \ 3 | \ |____\ -2√2 x-axis ``` Feel free to recreate this diagram on graph paper or using a drawing tool for a clearer visual understanding!