| Analytic Trigonometry |
| Lines a1 and a2 in two-space intersect at a point, forming an acute angle A. The slope of a1 is √3, and cos A = 1/7.
What is the slope of a2?
Please note: all answers must be in exact radical form. |
Problem Moderated by: Sasha |
| Problem Solution |
| We'll define the point of intersection of the two lines as our origin, which will make this problem a lot easier. The equations of our lines, then, follow:
{a1: y = x√3
{a2: y = mx,
where m is our unknown.
Call the origin A. On a1, the point B will be (1,√3), and the point C on a2 will be (1,m). From the information given, we determine the following:
c = 2
a = |m - √3|
b = √(1+m2)
We can quickly apply the law of cosines to solve for m:
a2 = b2 + c2 - 2bc cos A
(m - √3)2 = (√(1+m2))2 + 22 - 2(2)√(1+m2)(1/7), which reduces to a quadratic equation,
143m2 + 98m√3 + 45 = 0
In the process of deriving this equation, we have
m√3 + 1 = 2/7√(1+m2), so
m >= -1/√3
We'll use the quadratic formula to solve for m:
m = (-98√3 ± √(98*98*3 - 4*143*45)/(2*143) =
(-98√3 ± 32√3)/286 = -3√3/13 or -5√3/11
By inspection, we see that -5√3/11 < -1/√3. However, this implies that cos A = -1/7, which means that the supplementary angle of A which is also formed with this intersection, A', exists such that cos A' = 1/7 . Therefore, our two solutions are: -3√3/13 and -5√3/11 |
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