tag:blogger.com,1999:blog-5159574672638793285.post8035317115824115519..comments2024-01-29T06:01:15.858-05:00Comments on The Center of Math Blog: Problem of the WeekCenter of Mathematicshttp://www.blogger.com/profile/06584574038886207370noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5159574672638793285.post-9668203309736951842016-07-05T23:33:50.162-04:002016-07-05T23:33:50.162-04:00If you look at the congruent right triangles each ...If you look at the congruent right triangles each with the short side of length 1 which are generated by the radius of circle B along the x-axis and the radius along line m, you can write it as tan(1/2*(\pi - \theta)).<br />rephilhttps://www.blogger.com/profile/06634672785458640725noreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-83312657773284308062015-10-21T11:11:36.072-04:002015-10-21T11:11:36.072-04:00i solved it with direction and orthogonal vectors,...i solved it with direction and orthogonal vectors, linked with the equation of l.<br />same answer !maelnoreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-86536807283375597952015-08-31T09:40:02.953-04:002015-08-31T09:40:02.953-04:00I solved it via similar triangles, using the fact ...I solved it via similar triangles, using the fact that the hypotenuse of the smaller right triangle is 1/cos(theta) from SOHCAHTOA. It gets the same answer, of course. <br /><br />Your proof does not address what happens when theta is a right angle. Of course your formula gets the right result, but to be complete the proof should mention it. Anonymousnoreply@blogger.com