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Let θ be the angle between A and B. By the law of cosines,
|A+B|^2 = |A|^2 + |B|^2 - 2|A||B| cos θ
|A-B|^2 = |A|^2 + |B|^2 - 2|A||B| cos (π-θ) (I replaced |-B| with |B|).
The expressions in the first equation are 90^2 = 8100 times larger than the expressions in the second equation. Also note that cos θ = -cos (π-θ).
|A|^2 + |B|^2 - 2|A||B| cos θ = 8100(|A|^2 + |B|^2 + 2|A||B| cos θ)
Replace |B| with |A| and factor:
2|A|^2 (1 - cos θ) = 8100 (2|A|^2)(1 + cos θ)
Cancel out 2|A|^2 leaving
1 - cos θ = 8100(1 + cos θ) --> 8101 cos θ = -8099 --> cos θ = -8099/8101. To find θ just take arccos of both sides.
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The problem can also be solved by drawing a right triangle with sides 1, 90, sqrt(8101), drawing the median to the hypotenuse, and using area formulas, but it's a little tricky explaining without a diagram.
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