calculate sides of the right triangle if I know one side and all the angles
How to calculate the sides and hypotenuse length of the right triangle if I know $ \text{bigger side} = 60$, $\text{one angle} = 60^o$ & $\text{second angle} = 30^o$ ($\text{third angle} = 90^o$)
$\endgroup$4 Answers
$\begingroup$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
$$\cos(60)=\frac{adyacent-side}{hypotenuse} \Rightarrow adyacent-side = \cos(60)\cdot hypotenuse =\frac{60}{\sqrt{3}} $$
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 4 $\begingroup$Well, you don't even need to assume that the triangle is a right triangle. Simply use the Law of sines:
$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} .$$
$\endgroup$ $\begingroup$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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