MathDB

21

Part of 1986 AMC 12/AHSME

Problems(1)

Trigonometry

Source: AHSME 1986 problem 21

10/1/2011
In the configuration below, θ\theta is measured in radians, CC is the center of the circle, BCDBCD and ACEACE are line segments and ABAB is tangent to the circle at AA.
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(0,-1), E=(0,1), C=(0,0), D=dir(10), F=dir(190), B=(-1/sin(10*pi/180))*dir(10); fill(Arc((0,0),1,10,90)--C--D--cycle,mediumgray); fill(Arc((0,0),1,190,270)--B--F--cycle,mediumgray); draw(unitcircle); draw(A--B--D^^A--E); label("AA",A,S); label("BB",B,W); label("CC",C,SE); label("θ\theta",C,SW); label("DD",D,NE); label("EE",E,N); [/asy]
A necessary and sufficient condition for the equality of the two shaded areas, given 0<θ<π20 < \theta < \frac{\pi}{2}, is
<spanclass=′latex−bold′>(A)</span> tan⁡θ=θ<spanclass=′latex−bold′>(B)</span> tan⁡θ=2θ<spanclass=′latex−bold′>(C)</span> tan⁡θ=4θ<spanclass=′latex−bold′>(D)</span> tan⁡2θ=θ<spanclass=′latex−bold′>(E)</span> tan⁡θ2=θ <span class='latex-bold'>(A)</span>\ \tan \theta = \theta\qquad<span class='latex-bold'>(B)</span>\ \tan \theta = 2\theta\qquad<span class='latex-bold'>(C)</span>\ \tan \theta = 4\theta\qquad<span class='latex-bold'>(D)</span>\ \tan 2\theta = \theta\qquad \\ <span class='latex-bold'>(E)</span>\ \tan \frac{\theta}{2} = \theta
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