<p>i got 3.6, but I solved it using 2 steps, it was longer so here's the set up</p>
<p>tan(25)=2/x, where x is the lenth of the tangent line touching at the point, and x=about 4.28
then sin(25)=y/4.28, y is half the length of AB, and i got 3.62 for 2y</p>
<p>how do you solve #31? i kept getting 2.3 hrs, but the correct answer is 0.8</p>
<p>and #32..i'm kind of confused..</p>
<p>siyi, I see the validity in your method, but I can't see the error in my method.</p>
<p>oh!! thank you! comisar and il bandito
and i think you got the angle wrong, because 180-50=130, but in a triangle there's 3 angles, and i don't see how you got the 2 from..</p>
<p>Il bandito... I plugged your answer into my calculator and it came out 3.6. There was nothing wrong with your method, you just got the wrong answer somehow.</p>
<p>It's the funniest thing, my calc was on radian mode. I'll remember that on the test.</p>
<p>on the real sat II math 2c test can someone explain number 35 its like a triangle inscribed in a semicircle, and it asks for the area in terms of theta...</p>
<p>il bandito, how did you get 2, and 130? i can see it's in the form of the law of cosines..but i don't get how the tangent lines are 2</p>
<p>Siyi - they're radii of the circle. Also, because angle ACB is 50 degrees and each of the angles formed with the tangent lines is 90 degrees, 360 (sum of angles in a quadrilateral) - (50 + 90 + 90) = 130</p>
<p>Yeah 180n - 360 = angle sum in an n-gon. </p>
<p>Connect 2 radii from center to the two tangent points and you have a 4-gon, more specifically a kite. So then its 50-90-90-130. Use Law of Cosines to solve for the chord.</p>
<p>thx! i figured it out last night when i was trying to sleep..argh..I think i really need to think outside the box</p>
<p>guys wut about #21 in the booklet</p>
<p>and #29, in that same booklet, pdf i mean</p>
<p>nobody, c'mon guys, I neeeeeeeed u</p>