[imath]\bold{Challenge}[/imath] - looks simple but it's very difficult

\(\displaystyle c^2 = a^2 + b^2 = 5.2^2 + b^2\)
cal_3.png
\(\displaystyle c^2 = 27.04 + b^2\)
 
\(\displaystyle 46.4^{\circ} = 46.4^{\circ} \times \frac{\pi}{180^{\circ}} = 46.4^{\circ} \times \frac{3.14}{180^{\circ}} \ \text{rad}\)

challenge_3.png
\(\displaystyle 46.4^{\circ} = \frac{145.696}{180} \ \text{rad}\)
 
logistic_guy said:
Easy on me. It is a Challenge not \(\displaystyle 1 + 1\) kindergarten's problem. I need some time to solve it.

:cautious:😒
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I fail to see how this problem can be solved without using traditional trigonometric functions. I initially considered applying rational trigonometry, but that approach isn’t feasible when the given information includes angles without any associated coordinates. Since calculators are not allowed, are you planning to approximate the trigonometric values by hand using methods like the Taylor series?
 
logistic_guy said:
\(\displaystyle \sin 46.4^{\circ} = \frac{5.2}{c}\)
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Using Taylor series. Four our purpose, the first three terms are enough.

\(\displaystyle \sin x = x - \frac{x^3}{6} + \frac{x^5}{120}\)

Then,

\(\displaystyle \sin 46.4^{\circ} = \sin 0.809422 = 0.809422 - \frac{(0.809422)^3}{6} + \frac{(0.809422)^5}{120}\)
 
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