Problem 26 - Algebra 2 and Trigonometry

Question

For each \(\triangle O R S, O\) is the origin, \(R\) is on the positive ray of the \(x\) -axis and \(\overline{P S}\) is the altitude from \(S\) to \(\overrightarrow{O R}\) . Find the exact coordinates of \(R\) and \(S .\) b. Find the exact area of \(\triangle O R S .\) \(O R=8, \mathrm{m} \angle R O S=\frac{3 \pi}{4}, O S=8\)

Step-by-Step Solution

Answer

Coordinates of \(R\) are \((8,0)\), \(S\) are \((-4\sqrt{2}, 4\sqrt{2})\); Area is \(16\sqrt{2}\).

Step 1: Identify Triangle Components

In the given triangle, point \(O\) is the origin, so its coordinates are \((0, 0)\). The point \(R\) lies on the \(x\)-axis, and we are given that \(OR = 8\). This implies that the coordinates of \(R\) are \((8, 0)\) since it must lie on the \(x\)-axis.

Step 2: Find Coordinates of S

We are told that \(OS = 8\) and \( \angle ROS = \frac{3\pi}{4} \). The angle is measured counter-clockwise from the positive \(x\)-axis. To find \(S\), use polar coordinates: \((r, \theta)\), where \(r = OS = 8\) and \(\theta = \frac{3\pi}{4}\). Convert these to Cartesian coordinates using: \[ x = r \cos(\theta) = 8 \cos\left(\frac{3\pi}{4}\right) = 8 \left(-\frac{\sqrt{2}}{2}\right) = -4\sqrt{2} \] \[ y = r \sin(\theta) = 8 \sin\left(\frac{3\pi}{4}\right) = 8 \left(\frac{\sqrt{2}}{2}\right) = 4\sqrt{2} \] Thus, the coordinates of \(S\) are \((-4\sqrt{2}, 4\sqrt{2})\).

Step 3: Verify the Altitude Condition

\(\overline{PS}\) being the altitude implies that \(PS\) is perpendicular to \(\overrightarrow{OR}\). Since \(OR\) lies along the \(x\)-axis, \(PS\) must be a vertical line segment that does not change \(x\)-coordinate. Since \(S\) at \((-4\sqrt{2}, 4\sqrt{2})\) does provide a vertical line to some point on \(OR\), the conditions are met.\(P\) is not needed as it's an intermediary step to verify alignment.

Step 4: Determine the Area of \(\triangle ORS\)

The area \(A\) of a triangle can be found using the formula: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] For \(\triangle ORS\), the base \(OR = 8\) and the perpendicular height \(y\)-coordinate of \(S\), is \(4\sqrt{2}\) (as \(PS\) is vertical and aligns with \(y\)). Therefore:\[ A = \frac{1}{2} \times 8 \times 4\sqrt{2} = 16\sqrt{2} \]

Key Concepts

Cartesian Coordinates

Understanding the Cartesian coordinate system is key to navigating through many mathematical problems involving geometry. In this system, any point on a plane is represented by a pair of numbers called coordinates. These numbers specify the point in terms of its horizontal (x) and vertical (y) distances from a reference point known as the origin, denoted as (0, 0).
In the triangle \( riangle ORS\), point O is at the origin. This simplifies determining the coordinates of point R which lies on the x-axis. Since O is at (0, 0) and the distance OR is 8 units, R is simply at the point (8, 0). These are easy to visualize as movement along the x-axis without any change in the y-coordinate.
Using Cartesian coordinates allows us to break down problems into smaller, more manageable steps. It's like having a precise roadmap that tells you exactly where each point of interest is located.

Polar Coordinates

Polar coordinates provide a different way to describe the position of a point in the plane, using a combination of a distance from the origin and an angle from the positive x-axis. In contrast to Cartesian coordinates, which use (x, y), polar coordinates are expressed as (r, θ).
For point S in \( riangle ORS\), we've been given that the distance (r) from the origin O is 8, and the angle (θ) with the positive x-axis is \(\frac{3\pi}{4}\). To convert polar coordinates \(8, \frac{3\pi}{4}\) into Cartesian coordinates, we use the formulas:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)

Inserting the given values, we compute the coordinates of S as \(-4\sqrt{2}\) for x and \(+4\sqrt{2}\) for y, resulting in Cartesian coordinates of \((-4\sqrt{2}, 4\sqrt{2})\).
Polar coordinates are particularly useful in scenarios involving circles and angles, providing an intuitive understanding of how points relate to each other based on their direction and distance from a central point.

Triangle Area Calculation

Calculating the area of a triangle is a fundamental skill in geometry. For a triangle with a known base and height, the area can be easily determined with the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In the example of \( riangle ORS\), the base is the line segment OR, which measures 8 units. The height is determined by the y-coordinate of point S, which is \(+4\sqrt{2}\). Using these values in the formula gives:

• \(A = \frac{1}{2} \times 8 \times 4\sqrt{2} = 16\sqrt{2}\)

This result shows how effectively using the properties of triangles allows us to calculate their areas precisely. Understanding the orientation of triangles within coordinate planes simplifies the process significantly.

Algebra 2 Concepts

Algebra 2 expands upon the principles of basic algebra and is essential for solving geometric problems in coordinate geometry. Concepts from Algebra 2 like trigonometric identities, coordinate conversion, and simpler understanding of transformations can be crucial.
Finding polar coordinates and understanding their conversion to Cartesian coordinates requires the application of trigonometric functions from Algebra 2. When given an angle like \(\frac{3\pi}{4}\), being familiar with the sine and cosine functions allows us to find the precise x and y coordinates through:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)

By extending these concepts, students can easily interconnect different mathematical disciplines and apply them in various contexts from physics to computer graphics, showcasing the importance of a thorough understanding of these algebraic methods.