Math Problem Statement
Assume 𝜃 lies in quadrant 3 and the terminal side of 𝜃 is perpendicular to the line
y = −15 x + 3
Part 1: Determine sin(𝜃) Part 2: Determine sec(𝜃) sin(𝜃) = sec(𝜃) =
Solution
To solve this problem, we need to use the geometric relationships of the given line and the properties of angles in the third quadrant.
Part 1: Determine sin(𝜃)
The terminal side of the angle is perpendicular to the line . The slope of this line is -15. Since the terminal side is perpendicular to this line, the slope of the terminal side must be the negative reciprocal of -15, which is:
This gives us the rise over run for a point on the terminal side of the angle. Consider a right triangle formed by a point on the terminal side, the origin, and the projection onto the x-axis. The sides of the triangle will be:
- Opposite side = 1 (rise)
- Adjacent side = 15 (run)
We can now calculate the hypotenuse using the Pythagorean theorem:
In quadrant 3, both the -coordinate and the -coordinate are negative. Therefore, the opposite side is -1 and the adjacent side is -15.
Now, we can find :
So:
Part 2: Determine sec(𝜃)
Next, we find . Since , we first need to find .
Now, using this, we can determine :
Thus:
Final Answers:
Would you like further details or have any questions?
Here are 5 related questions to consider:
- How would the signs of trigonometric functions change in other quadrants?
- What is the reciprocal of each trigonometric function?
- How do we determine the cotangent given this scenario?
- Can you explain how to derive the cosine from the sine of an angle?
- How do you simplify expressions like ?
Tip: When solving trigonometric problems involving perpendicular lines, always start by finding the slopes and use Pythagorean relationships to handle the geometry.
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Math Problem Analysis
Mathematical Concepts
Trigonometry
Geometry
Slope of a Line
Formulas
sin(𝜃) = opposite / hypotenuse
sec(𝜃) = 1 / cos(𝜃)
cos(𝜃) = adjacent / hypotenuse
Pythagorean theorem: r = √(x^2 + y^2)
Theorems
Pythagorean Theorem
Trigonometric Functions in Different Quadrants
Suitable Grade Level
Grades 10-12
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