<p>This is the #35 question in the real sat II book, where they give you a picture of a triangle in a semicircle and then ask to express the area in terms of theta. The book says the answer is E, with no explanation. Could someone explain?</p>
<p>anyone?
i would love to write the problem... but its a drawing... :(</p>
<p>My method is a little imprecise, but it's the best I can do.</p>
<p>OK. Assume angle theta = 45 degrees. You can calculate the actual area of the triangle as being (1/2)b<em>h =(1/2)(1)(1)</em>2 = 1. </p>
<p>Now try the answer choices and see which one of them gives you 1 for the triangle's area. Answers C and E both give you 1, because tangent of 45 degrees = 1 and 2sin45*cos45 = 1 too. </p>
<p>But you can eliminate choice C, because tangent is the ratio of a triangle's opposite side to its adjacent side, and it is impossible that this is equal to the triangle's area for any angle theta. Therefore 45 must be a peculiar case where it works, but E must be the real answer.</p>
<p>Awkward I know, but it works.</p>
<p>using geometry you know that the triangle is a right triangle since it is an triangle inscribed in a semicircle. you know the hypotenuse of this triangle is 2. there are two ways to solve it from here:</p>
<p>1) using 1/2 * side1 * side2 * sin(theta) where the sides are the two sides that form theta.</p>
<p>you know one of the sides (2) to find the other side you can use trig:
2*cos(theta)= side 2</p>
<p>therefore:
.5*(2)(2cos(theta))(sin(theta))
=(2)(cos(theta))(sin(theta))</p>
<p>or
2) using .5<em>b</em>h</p>
<p>you know that the legs of a right triangle are the base and the heights, so you can use trig to find both legs:</p>
<p>2 sin(theta) and 2 cos (theta) are the legs</p>
<p>therefore:
.5<em>(2</em>cos(theta))(2*sin(theta))
=(2)(cos(theta))(sin(theta))</p>
<p>Thus the answer is E</p>
<p>-Aaron</p>