Line AB passes through points A(-6,6) and B(12,3). If the equation of the line in slope-intercept form,
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expand the binomial, either binomial theorem, or just FOIL
bear in mind, we're assuming the coefficient "a" is 1
and we're also assuming is the first form, it could be the second, but we're assuming is a vertical parabola
Well, the cos(∅) is negative in the second and third quadrants. If you are solving for theta, you would use the inverse of the cos or arccos
take the inverse of both sides to get:
x = arccos(-2/3) now evaluate the right
x = 131.8103149 degrees
to find your second solution, subtract your reference angle from 360 degrees.
360 - 131.8103149 = 228.1896851 degrees
Now the period of cos is 2π or 360 degrees. So if you want to consider all possible solutions, you would need to add/subtract 360n to both solutions above..
Not sure if this is what you're looking for, but thought I would leave it here for you just in case... As a side note, you could do this problem in radian measurement as well.
x = arccos(-2/3) now evaluate the right
x = 131.8103149 degrees
to find your second solution, subtract your reference angle from 360 degrees.
360 - 131.8103149 = 228.1896851 degrees
Now the period of cos is 2π or 360 degrees. So if you want to consider all possible solutions, you would need to add/subtract 360n to both solutions above..
Not sure if this is what you're looking for, but thought I would leave it here for you just in case... As a side note, you could do this problem in radian measurement as well.