This question is from my workbook (Pre-Calculus 11). Can someone please help me understand it?
Suppose theta is an angle in standard position with terminal army in quadrant III and tan theta = 6/3. Determine the exact calues of the other two primary trigonometric rations.
In quardatn III, x is ______ and y is ________ (r is always positive).
I don't understand how to draw the triangle for reference.
Any help will be greatly appreciated. Thank you.
(-6r)-5=7-12r
tan = y/x
since both x and y are negative in QIII, then you must have
tan = 6/3 = -6/-3,
so y = -6 and x = -3
now, r = 3sqrt(5)
sin = y/r
cos = x/r
just plug and chug
To understand how to draw the triangle in quadrant III, let's first recall what each trigonometric ratio represents in a right triangle.
In a right triangle, we have three sides: the hypotenuse (r), the adjacent side (x), and the opposite side (y). The primary trigonometric ratios are defined as follows:
- Sine (sin): sin(theta) = y/r.
- Cosine (cos): cos(theta) = x/r.
- Tangent (tan): tan(theta) = y/x.
In the given problem, we are told that tan(theta) = 6/3. Since tan(theta) = y/x, we can substitute the values and solve for y and x.
tan(theta) = y/x = 6/3
To simplify the equation, we can reduce 6/3 to 2/1:
2/1 = y/x
So, now we know that y = 2 and x = 1.
To draw the triangle in quadrant III, start by drawing the coordinate axes (x-axis and y-axis). In quadrant III, the x-values are negative (since they are to the left of the origin) and the y-values are also negative (since they are below the origin).
Place your triangle in quadrant III, with the origin (0, 0) at the vertex and the hypotenuse (r) in the quadrant III angle. The adjacent side (x) will be from the origin to the x-value (-1), and the opposite side (y) will be from the origin to the y-value (-2). Remember that the hypotenuse (r) is always positive, so draw it without any negative sign.
The triangle should look as follows:
|
r |
|
|
______|______
-1 -2
Now, we can find the values of the other two primary trigonometric ratios.
sine(theta) = sin(theta) = y/r = -2/r
cosine(theta) = cos(theta) = x/r = -1/r
Since r is always positive, we can simplify the answers as:
sine(theta) = -2/r
cosine(theta) = -1/r
So, the exact values of the other two primary trigonometric ratios in quadrant III are:
sine(theta) = -2/r
cosine(theta) = -1/r
I hope this helps you understand how to draw the triangle and determine the exact values of the other two primary trigonometric ratios. Let me know if you have any further questions!