Question

An angle x is chosen at random from the interval 0^0 < x < 90^0 Let p be the probability that the numbers sin^2 x, cos^2 x and sin x cos x are not the lengths of the sides of a triangle. Given that p = d/n where d is the number of degrees in arctan m and m and n are positive integers with m + n < 1000 find m + n.

Answer 1

Answer: 92

Explanation:

Observe that the probability is symmetric around
`45^(\circ)`.

If
`0^(\circ) < x < 45^(\circ)`, then
`\cos^2 x > \cos x > \sin x`. By the triangle inequality, it follows that
`\cos^2 x > \sin^2 x+\sin x \cos x`.

We can now rearrange as follows:

`\cos^2 x > \sin^2 x+\sin x \cos x\\\\\cos^2 x -\sin^2 x > \sin x \cos x\\\\\cos 2x > (1)/(2)\sin 2x`

Since
`\cos 2x` and
`\sin 2x` are both positive for the chosen interval,

`2 > \tan x \implies x < (1)/(2)\arctan 2`.

Therefore, the probability is
`((1)/(2) \arctan 2)/(45)=(\arctan 2)/(90)`.

This means,
`m=2, n=90 \implies m+n=92`.