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

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

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Answer 1 · 2015-06-25T02:54:20

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

Explanation:

Both $30^{o}$ $30^o$ and $60^{o}$ $60^o$ are based on an equilateral triangle with sides of length 2 and with one of the angles bisected.

The $45^{o}$ $45^o$ 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
$XXXX$ $color(white)("XXXX")$ sin $= \frac{opposite}{hypotenuse}$ $= "opposite"/"hypotenuse"$

$XXXX$ $color(white)("XXXX")$ cos $= \frac{adjacent}{hypotenuse}$ $= "adjacent"/"hypotenuse"$

$XXXX$ $color(white)("XXXX")$ tan $= \frac{opposite}{adjacent}$ $= "opposite"/"adjacent"$

and their reciprocals.

Then, for example:
$XXXX$ $color(white)("XXXX")$ $sin (60^{o}) = \frac{\sqrt{3}}{2}$ $sin(60^o) = sqrt(3)/2$

$XXXX$ $color(white)("XXXX")$ $sin (30^{o}) = \frac{1}{2}$ $sin(30^o) = 1/2$

$XXXX$ $color(white)("XXXX")$ $sin (45^{o}) = \frac{1}{\sqrt{2}}$ $sin(45^o) = 1/sqrt(2)$

$XXXX$ $color(white)("XXXX")$ $cos (60^{o}) = \frac{1}{2}$ $cos(60^o) = 1/2$

etc.

Answer 2 · 2018-06-04T13:27:55

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.

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

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