the bottom or base edge b,the top left edge a,the top right edge c.Let a=c+2 and b=10.

Question

Isla Klein · Accepted Answer

Taking the origin as the intersection of a and b, with the equation in terms of the angle

θ

at that vertex:
Through the cosine law we have:

(a - 2)^{2} = a^{2} + 100 - 20 a \cos θ

- 4 a + 4 = 100 - 20 a \cos θ

a = \frac{96}{20 \cos θ - 4}

With our particular setup, a is the radius. So this is a polar equation, where

r (θ) = \frac{96}{20 \cos θ - 4} = \frac{24}{5 \cos θ - 1}

we can convert to cartesian coordinates if you want:

5 r \cos θ - r = 24

(5 r \cos θ - 24)^{2} = r^{2}

(5 x - 24)^{2} = x^{2} + y^{2}

y^{2} = 24 x^{2} - 240 x + 576

y^{2} = 24 (x - 5)^{2} - 24

(x - 5)^{2} - \frac{y^{2}}{24} = 1

which is a hyperbola. You want the rightmost branch (so

a > 0

), which is valid when

\cos θ < \frac{1}{5}

the bottom or base edge b, the top left edge a, the