How do you find the exact values #cos (pi/6)# using the special triangles?

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Answer 1 · 2018-07-15T02:54:55

As below.

![https://www.dummies.com/education/math/calculus/how-to-work-with-30-60-90-degree-triangles/]( useruploads.socratic.org )

#(pi/6)^c = (pi/6) * (180/pi) = 30^@#

#cos (pi/6)^c = cos (30)^@#

#"Angles are " 30^@, 60^@, 90^@ " and the sides are "a, a sqrt3, 2a#

#cos theta = " Adj. side / Hypotenuse" = (AB) / (AC)#

#cos 30 = (cancela sqrt3) / (2 cancela) = sqrt3 / 2#