ASKSAGE: Sage Q&A Forum - Individual question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Fri, 22 Oct 2010 12:49:00 -0500SAT Math Level 2 Subject Test Sample Question #26http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/A circle has a radius of 2. Two different tangent lines intersect the circle at points D and E. Those same tangent lines intersect each other outside the circle at point F, forming an acute angle of 50 degrees. If a chord is drawn between D and E, what is its length?Wed, 20 Oct 2010 05:18:26 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/Answer by Mike Hansen for <p>A circle has a radius of 2. Two different tangent lines intersect the circle at points D and E. Those same tangent lines intersect each other outside the circle at point F, forming an acute angle of 50 degrees. If a chord is drawn between D and E, what is its length?</p>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?answer=11749#post-id-11749This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is `180 - 50 = 130` degrees. Half of the chord DE is a side of a right triangle opposite an angle of `130 / 2 = 65` degrees. Using the formula for `sin` and doubling we get the chord length to be
sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
Wed, 20 Oct 2010 09:40:22 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?answer=11749#post-id-11749Comment by ccanonc for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22545#post-id-22545Nope, it's an SAT question. And the questions were slowing down, so it bumped the traffic up a bit...and the exercise sometimes brings up other issues. Are you saying sage shouldn't be used for teaching, or problems you don't like shouldn't be asked?Wed, 20 Oct 2010 10:08:43 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22545#post-id-22545Comment by kcrisman for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22541#post-id-22541For instance, on a Geogebra site, this question would be great, because you can construct the items in question. The only part Sage can do here is approximate.
Some GRE question asking people to demonstrate the Cayley tables of two non-isomorphic groups of order four would be perfect, though!Wed, 20 Oct 2010 15:58:03 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22541#post-id-22541Comment by cswiercz for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22540#post-id-22540I second Mike's comment. Especially in light of the recent flood or GRE and SAT problems posted on the discussion board. Sure, some of these problems make great examples of Sage in-use but the theoretical ones are more appropriate for a math help forum.Wed, 20 Oct 2010 16:23:29 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22540#post-id-22540Comment by ccanonc for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22539#post-id-22539Two a day is not a flood, especially when the traffic is dead. (I hear Mel Gibson crying "FREEDOM!";) The time it took you to complain, you could have answered the question and demo'ed sage. The ratio of squeaky wheels to grease is high. I'd rather be flogged than listen to this.Wed, 20 Oct 2010 17:06:18 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22539#post-id-22539Comment by kcrisman for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22542#post-id-22542I think that Mike's point is just that ask.sagemath.org can't be all things to all people. Until it gets a little more off the ground, it's wise to make it focused; otherwise it's confusing as to what it is.
That said, if Sage really brings insight to an SAT or GRE problem, bring it on!Wed, 20 Oct 2010 15:54:57 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22542#post-id-22542Comment by Mike Hansen for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22544#post-id-22544Sage should be used for teaching, and this website should be used for teaching about Sage. But, I don't think this website should really be used for teaching math or solving homework problems. While it may have bumped the traffic up, the question and answer probably weren't useful to anyone.Wed, 20 Oct 2010 11:27:25 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22544#post-id-22544Comment by Evgeny for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22538#post-id-22538It is not easy to create a successful community site. To make this happen you first of all should use it for your own benefit. If you make it useful for yourself and have patience - then others will join. Secondly, you should think of how to make this site better. e.g.- why not make a better logo?Fri, 22 Oct 2010 12:49:00 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22538#post-id-22538Comment by ccanonc for <p>This isn't really a Sage question. The angle between the center of the circle C, point D (or E), and F is a right angle. Thus, the angle DCE is <code>180 - 50 = 130</code> degrees. Half of the chord DE is a side of a right triangle opposite an angle of <code>130 / 2 = 65</code> degrees. Using the formula for <code>sin</code> and doubling we get the chord length to be</p>
<pre><code>sage: 2*(2*sin(65/360*2*pi))
4*sin(13/36*pi)
sage: _.n()
3.62523114814660
</code></pre>
http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22543#post-id-22543It does teach about sage, but you're the admin, so if you want to pull rank, just delete my post. Vive le Censorship.Wed, 20 Oct 2010 12:09:26 -0500http://ask.sagemath.org/question/7737/sat-math-level-2-subject-test-sample-question-26/?comment=22543#post-id-22543