0
$\begingroup$

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

4 Answers 4

1

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)

  • 0
    how is this going to help me calculate the side and the hypotenuse?2010-11-29
3

You could use the identities

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

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

In fact you have :

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

and

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

  • 0
    @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 triangle2010-11-29
2

Some hints: what is the third angle? Draw a picture. Can you spot some symmetry? If not, then decode the following hint with http://www.rot13.com/index.php. 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.

  • 0
    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.2010-11-29
2

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

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