How do you sketch the angle whose terminal side in standard position passes through #(1,sqrt3)# and how do you find sin and cos?

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Feb 9, 2018


#sin(x) = y/(Hyp) = sqrt(3)/2# and #cos(x)=x/(Hyp) = 1/2#


If you place the triangle on a graph, with one side laying on the positive side of the x-axis, and set the other side to pass through the origin and the given point #(1, sqrt(3))# it will form a right triangle.

The bottom side will be the "adjacent" side to the angle, while the vertical side is the "opposite."

knowing the acronym SOH CAH TOA
#sin=(Opposite)/(Hypotenuse)# , #cos=(Adjacent)/(Hypotenuse)# , #tan=(Opposite)/(Adjacent)#

and the Pythagorean Theorem
#(Hypotenuse)^2 = (Opposite)^2 + (Adjacent)^2#

We can find that
#Hyp^2 = (sqrt(3))^2 + 1^2#
By solving for Hyp, we see that the hypotenuse is 2 units long.

Knowing this, we can then solve the sine and cosine.

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