To find m<2 when m<1 and m<2 are given as vertical angles, set the equations equal: 17x + 1 = 20x - 14. Solving for x yields x = 5. Substituting x back into either angle's equation, m<1 or m<2, results in m<2 = 101. Therefore, the measure of angle 2 is 101 degrees.
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bernardl
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<1 and <2 are vertical angles. If m<1 = 17x + 1 and m<2 = 20x - 14, find m<2.
As vertical angles have the same measure, set 17x + 1 = 20x - 14, solve for x and substitute that value for x into the expression for angle 2 (or angle 1 if you prefer; they both have the same measure). The result is the desired measure.
I just saw this one. If there are finitely many primes, then
##0<\prod_{p}\sin(\frac\pi p)=\prod_p\sin\left(\frac{\pi(1+2\prod_q q)}p\right)=0##
Of course it is in a way just a variation of Euclid's idea, but it is a one liner.