How do you solve this problem?

Answer:
Step-by-step explanation:
Triangle problems, and right triangle problems in particular, can often be solved using trigonometry. The ratios of the different sides of a right triangle define different trig functions of the angles. The mnemonic SOH CAH TOA is often taught as a way to remember how the basic trig functions are defined. It means ...
Sin(angle) = Opposite/Hypotenuse
Cos(angle) = Adjacent/Hypotenuse
Tan(angle) = Opposite/Adjacent
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In this problem, you are given an angle and the measure of its opposite side. This means you can use the tangent function to find the adjacent side, and the sine function to find the hypotenuse. Of course, since this is a right triangle, knowing any two sides, you can use the Pythagorean theorem to find the third one, if you choose.
If you've been studying trigonometry for even a short while, you have probably been asked to memorize a short table of trig function values. Commonly, that table includes ...
sin(30°) = cos(60°) = 1/2
sin(45°) = cos(45°) = 1/√2 = (√2)/2
sin(60°) = cos(30°) = (√3)/2
Since tan(angle) = sin(angle)/cos(angle), the corresponding tangent values are ...
tan(30°) = 1/√3 = (√3)/3
tan(45°) = 1
tan(60°) = 1/tan(30°) = √3
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Solving for m first, we have ...
tan(60°) = 5/m
m = 5/tan(60°) = 5/√3 = (5/3)√3 . . . . . multiply by m/tan(60°) to find m
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Then solving for n, we have ...
sin(60°) = 5/n
n = 5/sin(60°) = 5/((√3)/2) = 10/√3 = (10/3)√3
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Comment on exact solutions
Most trig functions of most angles are irrational numbers, not necessarily "algebraic" numbers (roots of polynomials). Trig functions of 15° and powers of 2 of that angle can be written in exact form (perhaps involving several layers of square roots), but functions of most other angles cannot.