Problem 29 - 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 \(\triangle O R S .\) \(O R=7, \mathrm{m} \angle R O S=\frac{\pi}{6}, P S=8\)

Step-by-Step Solution

Answer

Coordinates: \( R = (7, 0), S = (4\sqrt{3}, 8) \). Area: 28.

Step 1: Determine coordinates of R

Point \( R \) is on the positive \( x \)-axis. Given that \( OR = 7 \) and \( O \) is the origin, the coordinates of \( R \) must be \( (7, 0) \).

Step 2: Use trigonometry to find S's coordinates

Given \( \angle ROS = \frac{\pi}{6} \) and \( PS = 8 \), we know that \( \triangle OPS \) is a 30-60-90 triangle. In such triangles, the hypotenuse is \( 2 \) times the shorter leg (which is the altitude \( PS = 8 \)). Using this, we find the opposite side (height) is \( 8 \), and using the 30-degree angle, the length \( OS = 16 \) is found using \( \sin(30°) = 1/2 \). So, the coordinates of \( S \) based on angle \( \angle ROS \) and distance are \( S = (16 \cos(\frac{\pi}{6}), 16 \sin(\frac{\pi}{6})) = (4 \sqrt{3}, 8) \).

Step 3: Calculate area of \(\triangle ORS\)

The area of a triangle can be calculated using the formula: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). Here, the base \( OR = 7 \) and the height \( PS = 8 \). Thus, the area of \( \triangle ORS \) is \( \frac{1}{2} \times 7 \times 8 = 28 \).

Key Concepts

Coordinates

Coordinates are like an address for points in a plane. They help us locate each point precisely. In a two-dimensional space, we use the x and y axes. The origin, marked as point \(O\), is the central reference and is located at \((0, 0)\).
For any point on the x-axis, like our point \(R\), the y-coordinate is always zero. Hence, it is crucial to recognize that if a point is just on the x-axis, its position can be expressed as \((x, 0)\).
In our problem, since \(R\) is getting its position from the origin and lies on the positive x-axis exactly 7 units away, the coordinates of \(R\) are \((7, 0)\). This also helps us when determining other points, such as \(S\), which requires more trigonometric analysis to locate precisely.

30-60-90 triangle

A 30-60-90 triangle is a special type of right triangle that has angles of 30 degrees, 60 degrees, and 90 degrees. The sides of this triangle have specific ratios that can help solve many geometry problems.
The ratio of the sides of a 30-60-90 triangle is always 1 : \(\sqrt{3}\) : 2. This means:

• The shortest side, opposite the 30° angle, measures \(x\).
• The side opposite the 60° angle, the middle length, measures \(x\sqrt{3}\).
• The hypotenuse, the longest side, measures \(2x\).

In our exercise, since \(PS\) (the shorter leg) measures 8, we use this ratio to find \(OS\). Here, \(OS = 2 \times 8 = 16\), which also gives the clue about where \(S\) is located when considering the precise coordinates in relation to \(O\) and \(R\).

Triangle Area

Calculating the area of a triangle requires understanding the formula which is: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\). This formula is simple but powerful, providing the solution as long as the base and height are known.
In the given exercise, \(OR\) serves as the base while \(PS\) is considered the perpendicular height from \(S\) to the x-axis (which aligns with \(OR\)). With \(OR = 7\) and \(PS = 8\), the area calculation proceeds as follows:
\[\text{Area} = \frac{1}{2} \times 7 \times 8 = 28\]
This calculation highlights the efficiency of using the area formula by substituting the base and height directly comparable to geometric understanding.

Trigonometrical functions

Trigonometrical functions are vital in transforming angles and distances into concrete coordinates on a plane. Sine, cosine, and tangent are core functions.
In this task, determining the exact location of point \(S\) involves using these functions. With given angle \(\angle ROS = \frac{\pi}{6}\) (or 30 degrees), the functions assist in calculating the missing sides and hence coordinates.
The key trigonometric equations are:

• For the x-coordinate of \(S\): \(S_x = OS \cdot \cos(\frac{\pi}{6}) = 16 \cdot \frac{\sqrt{3}}{2} = 8\sqrt{3} \).
• For the y-coordinate of \(S\): \(S_y = OS \cdot \sin(\frac{\pi}{6}) = 16 \cdot \frac{1}{2} = 8\).

By breaking the problem into these trigonometric components, we not only understand the coordinates but also enhance our skills in placing points using angles and known distances.