How do you find trigonometric ratios of 30, 45, and 60 degrees?

2 Answers
Jun 25, 2015

The trigonometric ratios for 30^o30o, 45^o45o, and 60^o60o are based on some standard triangles. sin, cos, and tan (and their reciprocals) are the ratios of the sides of these triangles.

Explanation:

Both 30^o30o and 60^o60o are based on an equilateral triangle with sides of length 2 and with one of the angles bisected.

The 45^o45o angle is based on an isosceles triangle with the equal sides having a length of 1.

For all triangles the Pythagorean Theorem is used to compute the "missing" side length.
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If remember that
color(white)("XXXX")XXXXsin = "opposite"/"hypotenuse"=oppositehypotenuse

color(white)("XXXX")XXXXcos = "adjacent"/"hypotenuse"=adjacenthypotenuse

color(white)("XXXX")XXXXtan = "opposite"/"adjacent"=oppositeadjacent

and their reciprocals.

Then, for example:
color(white)("XXXX")XXXXsin(60^o) = sqrt(3)/2sin(60o)=32

color(white)("XXXX")XXXXsin(30^o) = 1/2sin(30o)=12

color(white)("XXXX")XXXXsin(45^o) = 1/sqrt(2)sin(45o)=12

color(white)("XXXX")XXXXcos(60^o) = 1/2cos(60o)=12

etc.

Jun 4, 2018

The other answer is fine. I'll just point out that once the student has internalized these Two Tired Triangles of Trig (30/60/90 and 45/45/90) they're done. If you review the trig questions here you'll find the vast majority use just these triangles.