Could I find the two missing angle measures if I know some of the side lengths of a right triangle?

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Could I find the two missing angle measures if I know some of the side lengths of a right triangle?

Mathematics
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some means more than 1, right?
Yeah, I'm pretty sure
then with sine, cosine, tangent, you can find the angles

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yes, by using SAS congruents and those stuff. like.. a^2 = b^2 + c^2 -2ab Cos A
im not really sure how to use them to find the angles instead of the other way around
a^2 = b^2 + c^2 -2ab Cos A
But if you know it's a right triangle, you don't use the cosine law
Better only use a² + b² = c²
Got it?
that'll help me find the sides, but i need the angles
sine theta = opposite/ hypotenuse cosine theta = adjacent / hypotenuse tangent theta = opposite/ adjacent
Pythagorean theorem does not help to find angles
Yeah
so how do i use the trigonometric ratios to find the angles if i only know the right angle?
im still confused xD
then you use inverse sine/cosine/tangent
okay, how do i use those? im a bit new at this
The sum of the interior angles of a triangle are equal to 180 degree. To find the third angle of a triangle when the other two angles are known subtract the number of degrees in the other two angles from 180 degree :)
i understand that, but i only know the 90 degree angle
Let me show you the example so that you will get it better. :)
@sweetrascal That has nothing to do with this
Example: How many degrees are in the third angle of a triangle whose other two angles are 40degree and 65degree ? Answer: 180degree - 40degree -65degree = 75degree
i only know the 90 degree angle... that's not what im looking for
\[\sin \theta = \frac{1}{2}\]\[\sin^{-1}\sin \theta = \sin^{-1} \frac{1}{2} \implies \theta = \sin^{-1} \frac{1}{2}\]
Moneybird will help you further :)
hmm
so both sides are multiplied by the inverse sine?
not multiplied on your calculator there is a inverse sine buton
okay, i get it! thank you!

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