manually compute opposite and adjacent lengths when hypotenuse and angles are known in right triangle

Need to manually compute the opposite and adjacent lengths in a right triangle when all else is known.

IF using sine then must compute sine mathematically (calculator with no sin/cos functions allowed); also can't use pi.

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you could also start with sin and cos of a right angle, and use

sin(x/2) = sqrt((1-cos(x))/2)
cos(x/2) = sqrt(1+cos(x))/2)
sin(a+b) = sin(a)cos(b)+cos(a)sin(b)
cos(a+b) = cos(a)cos(b)-sin(a)sin(b)
until you get sufficiently close
sin is x - x^3/3! + x^5/5! - ...
cos is 1 - x^2/2! + x^4/4! - ...
BaltarAuthor Commented:
Using these formulas how far out do you have to go before you begin to approach the known values.

If x is in degrees I was way off initially and the trend towards the known value was going to be a very long time.

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x is in radians not degrees.
how far to go depends on how close you want to "approach":
I find it hard to believe that a question like that in general would be given in a test.
Can you give an example question complete with values?
BaltarAuthor Commented:
radians violates my no pi issue

this is not for a test, more of a personal growth project and an excuse to use a very powerfull computer to figure something out to thousands of decimal places

pi in most forms is actually not accurate enough so I need the formulas behind my right triangle question that do not use pi directly or indirectly (radians)

Try ozo's suggestions
radians do not have to contain pi directly or indirectly.
(pi is known to many thousand of decimal places which ought to be close enough for most purposes.)
>>when all else is known<<
Out of curiosity, what exactly is "all else"?
With standard labels, you seem to know alpha and c, hence beta = 90°-alpha and want to knw a, and b.
Whatever you do to compute a andb, you can compute sin(alpha) at just very little extra cost (namely by sin(alpha)=a/c)
Hence it is "impossible" to not use a calculation of sin(alpha).
The addition and half-angle formulas as given by ozo allow you to obtain all angles of the form n*15/2^k degrees with +,-,*,/ and square roots, but the other mothods described are probably much faster for the same high precision.
some sine and cosine of some angles, like pi/2 or pi/3 or pi/5 are easier to compute than others,
but any method that can compute it for arbitrary angles can be converted into a method for computing pi, so you can't really avoid using it in some manner
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