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The question says it all.

Given a triangle, find its angles without a calculator. Is this even possible without tables or making tables?


Is it possible to determine the inverse sin, cos of a triangle with an equation not involving calculators and tables?

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You will need a calculator or a table. Definitely. – ncmathsadist Mar 8 '13 at 2:51
up vote 1 down vote accepted

Use the Law of Cosines. If $C$ is the measure of the angle opposite the side of length $c$, and $a,b$ the other two side lengths, then $$c^2=a^2+b^2-2ab\cos C,$$ so $$\cos C=\frac{a^2+b^2-c^2}{2ab},$$ and so since $C$ is an angle of a triangle, then $$C=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right).$$

Similarly, if $A$ and $B$ are the respective measures of the angles opposite the sides of length $a$ and $b$, then $$A=\cos^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)$$ and $$B=\cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right).$$

If you want a numerical value, you're probably out of luck, but the above work does give us exact and correct values for $A,B,C$.

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So the answer doesn't exist without a means for obtaining cos^-1? Thx for the answer! – Corona Salad Mar 8 '13 at 2:13
The answers exist, and are given above. The numerical approximations to the answers will usually not be obtainable, though--or at least, not easily obtainable without a calculator or table. – Cameron Buie Mar 8 '13 at 2:15
@CoronaSalad In fact, your problem is equivalent to calculating $\cos^{-1}$ in the sense that any method that allows you to calculate $\cos^{-1}$ allows you to solve your problem and vice versa. So the two problems are "equally difficult". – Jack M Dec 31 '14 at 17:04

protected by Community Jul 3 '15 at 23:24

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