Question 1198656
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I'll use a geometric approach, and a bit of algebra.


What we can do is form this rectangle here:
*[illustration Screenshot_171.png]
Note the 15+45+30 = 90 in the top right corner.


The three triangles at the bottom are your standard template 30-60-90 triangles, and a 45-45-90 triangle.

We'll focus on the triangle up top. It has an angle of 15 degrees.
The tangent of this angle is the opposite over adjacent.


{{{tan(angle) = (opposite)/(adjacent)}}}


{{{tan(15^o) = (sqrt(3)-1)/(sqrt(3)+1)}}}


{{{tan(15^o) = ((sqrt(3)-1)*A)/((sqrt(3)+1)*A)}}} Multiply top and bottom by some variable, which I'll call A.


{{{tan(15^o) = ((sqrt(3)-1)(sqrt(3)-1))/((sqrt(3)+1)(sqrt(3)-1))}}} Plug in {{{A = sqrt(3)-1}}} so that the denominator is rationalized.


{{{tan(15^o) = ((sqrt(3))^2-2*sqrt(3)+(1)^2)/((sqrt(3))^2-1^2)}}} Expand the numerator. Use the difference of squares rule in the denominator.


{{{tan(15^o) = (3-2*sqrt(3)+1)/(3-1)}}} The square root is canceled out in the denominator.


{{{tan(15^o) = (4-2*sqrt(3))/(2)}}}


{{{tan(15^o) = (2(2-sqrt(3)))/(2)}}}


{{{tan(15^o) = 2-sqrt(3)}}}


Answer: Choice E
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