[johnak]

Lines a1 and a2 in two-dimensional space intersect at a point, forming an acute angle A. The slope of a1 is the square root of 3, and cosA=1/7. What is the slope of a2? There are two solutions.

[reinsuranc]

I'll set it up: you can finish it. This was not easy.


Let line a1 be y=(squareroot3)(x).
Let line a2 be y=mx, with m unknown, such that
the lines intersect at (0,0) making an acute angle A.

Since cosA=1/7, go a distance of 7 from (0,0) along a1, and call the
point a distance of 7 from (0,0) B.

Similarly, go a distance 1 from (0,0) along a2, and call the
point a distance of 1 from (0,0) C.

Now (0,0), B, and C form a right triangle with hypotenuse 7,
adjacent side to angle A with length 1, and an unknown third
side connecting B to C. By Pythagorean Theorem, the third
side has length (4)(squareoot3).

The coordinates of B are (x, (squareroot3)(x) ), but it is
a distance 7 from (0,0). Using the formula for the distance d
between 2 points (x1,y1) and (x2,y2) satisfies
d^2 = (x2-x1)^2 + (y2-y1)^2, where "^" indicates exponentiation,
7^2 = (x2-0)^2 + ((squareroot3)(x2)-0)^2, gives x2=plus or minus 7/2,
and y2=plus or minus (squareroot3)(7/2). So the
coordinates of B are plus or minus (7/2, (squareroot3)(7/2) ).

I am going to proceed with the plus part of this.

The coordinates of C are (x,mx). We just solved for the coordinates
of B. We know the length of side BC is (4)(squareoot3).
Again using the formula for the distance d between 2 points, we have
(4squareoot3)^2 = (7/2 - x)^2 + ((squareroot3)(7/2) - mx)^2.
We also know that by dropping a perpendicular from C to the x axis
we have x^2 + (mx)^2 =1. This can be solved for m and substituted
into (4squareoot3)^2 = (7/2 - x)^2 + ((squareroot3)(7/2) - mx)^2
to solve for x. Substitute the x answer back into
x^2 + (mx)^2 =1, and we are done.