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How to calculate the sides and hypotenuse length of the right triangle if I know $ \text{bigger side} = 60$, $\text{one angle} = 60^o$ & $\text{second angle} = 30^o$ ($\text{third angle} = 90^o$)

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up vote 2 down vote accepted

Some hints: what is the third angle? Draw a picture. Can you spot some symmetry? If not, then decode the following hint with But please, spend some time trying first.

Ersyrpg gur gevnatyr va gur evtug natyr. Jung vf fb fcrpvny va gur erfhygvat gevnatyr? Gel znxvat hfr bs gur rkgen flzzrgel.

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rtel the third is 90 because it's a right triangle – Omu Nov 29 '10 at 14:44
Exactly. Now stare at the picture for at least 20 minutes :-) Try things with the triangle. If you then still cannot do it, decode my hint. But remember - it's the 20 minutes that you spend before that, that make the exercise worthwhile. Feel free to ask again, if the hint is not sufficient. – Alex B. Nov 29 '10 at 14:47
rtel I found that one side is 2x times smaller than the hypotenuse, but this will work only when I have one angle of 60 degrees, I want to know what to do when I got one angle of 57 degrees for example – Omu Nov 29 '10 at 14:50
That's a different story, then you need trigonometry, namely the sine rule: the ratio between the sine of an angle and the length of the opposite side is the same for all the angles in the triangle. But what you have just said is exactly the right way to solve the concrete problem you have posted. – Alex B. Nov 29 '10 at 14:52

You could use the identities

$$sin(\theta)=\frac{oposite-side}{hypotenuse} $$

$$cos(\theta)=\frac{adyacent-side}{hypotenuse} $$

In fact you have :

$$\sin(60)=\frac{60}{hypotenuse} \Rightarrow hypotenuse = \frac{120}{\sin(60)}=\frac{120}{\sqrt{3}}$$


$$\cos(60)=\frac{adyacent-side}{hypotenuse} \Rightarrow adyacent-side = \cos(60)\cdot hypotenuse =\frac{60}{\sqrt{3}} $$

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Yocks and how is this going to help me calculate the side and the hypotenuse? – Omu Nov 29 '10 at 14:51
@Omu Draw the triangle on paper and compared to these calculations. Is essential that is a right triangle, but you can apply the law of sines if it is not right triangle – Bryan Yocks Nov 29 '10 at 15:02

Well, you don't even need to assume that the triangle is a right triangle. Simply use the Law of sines:

$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} .$$

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tan(60) = side1/side2 = 60/side2 => side2 = 60/tan(60) = 60/sqrt(3) = 20sqrt(3)

sin(60) = side1/hypotenuse => hypotenuse = 60/sin(60)=120/sqrt(3) = 40sqrt(3)

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how is this going to help me calculate the side and the hypotenuse? – Omu Nov 29 '10 at 14:52

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