Is there a way to manually find the angles of the $3:4:5$ triangle?
Been thinking of a solution all day. Can't find one that would satisfy.
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4The cosine rule? – Peter Foreman Dec 01 '19 at 21:41
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1without actual trigonometry, no – Saketh Malyala Dec 01 '19 at 21:43
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3What do you mean by 'manually'? – Jam Dec 01 '19 at 21:49
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The acute angles in that triangle are not nice well known multiples of $pi$. Trigonometry is all you have for finding the angles. – Ethan Bolker Dec 01 '19 at 21:50
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1You can 'manually' construct the triangle ... – Hagen von Eitzen Dec 01 '19 at 21:51
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I suppose it could go something like "We know one angle is $\pi/2$, so now we just have to show that the other angles have a ratio of $1/2$ or $2$ between them. There might be a geometric proof there? – QC_QAOA Dec 01 '19 at 21:52
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@QC_QAOA Your comment inspired me to post a question: (Question 3459004). I believe this shows that the not only is the ratio of the angles irrational but it is also transcendental. – Jam Dec 01 '19 at 23:54
2 Answers
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The best you can do is a decimal approximation. One way to do this by hand is to use the Taylor series for $\arctan x$. The first 5 terms are
$$\arctan(x) \approx x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}.$$
One angle is given by
$$\arctan(3/4) \approx 3/4-\frac{(3/4)^3}{3}+\frac{(3/4)^5}{5}-\frac{(3/4)^7}{7}+\frac{(3/4)^9}{9}$$
$$ = \frac{5928081}{9175040} = .6461095537,$$
which is not far from the actual value of $0.6435011088.$ If you want more accuracy, use more terms.
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Approximate the angle $\theta = \sin^{-1}\frac35$ via
$$\theta = \sin^{-1} t + (\sin^{-1} t)'\left(\frac35-t\right)$$
Let $t=\frac12$ and use $ \sin^{-1}\frac12 = \frac\pi6$, $(\sin^{-1} t)' = \frac1{\sqrt{1-t^2}}$ to compute $\theta$ as,
$$\theta = \frac\pi6 + \frac2{\sqrt3}\cdot\left(\frac35-\frac12\right) =\frac\pi6+ \frac{\sqrt3}{15}$$
which is 36.6$^\circ$ vs. the exact angle 36.8$^\circ$.
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