jlh@5941ux.UUCP (10/10/83)
To restate the isosceles triangle problem submitted some time ago, Given an isosceles triangle with angle A = 20 degrees angle B = angle C = 80 degrees, construct segment AD along AB such that length AD = length BC. Find angle ADC. ___________________________________________________________________ Well, I couldn't find a solution in under five minutes, but I did use some basic trigonometry to solve for the angle in about 10. It turns out that the angle is exactly 150 degrees, a surprising result!! Further investigation leads to the discovery that these numbers work sin 10 sin 20 out nicely (not unlike Pythagorean triples) because ------ = ------ . sin 30 sin 80 This identity does not seem intuitively obvious to me at all, but can be proven rather easily as follows. sin(80)sin(10) = 0.5*[cos(80-10)-cos(80+10)] = 0.5*cos(70) = 0.5*cos(90-20) = 0.5*cos(90)cos(20) + 0.5*sin(90)sin(20) = 0.5*sin(20) = sin(30)sin(20). sin 20 sin 40 sin 30 sin 50 Similarly, ------ = ------ and ------ = ------ . sin 30 sin 70 sin 40 sin 80 Did everyone out there know about these identities? Maybe I'm weird or maybe I'm naive, but this discovery (for me) was something akin to the feeling that came over me when I first discovered Pythagorean triangles. Jeff Heatwole ..!houxm!5941ux!jlh
rlh@mit-eddie.UUCP (Roger L. Hale) (10/19/83)
To restate the isosceles triangle problem submitted some time ago, Given an isosceles triangle with angle A = 20 degrees angle B = angle C = 80 degrees, construct segment AD along AB such that length AD = length BC. Find angle ADC. ___________________________________________________________________ One way comes to mind which worked for me in less than five minutes (suggested by past playing with cyclotomic numbers, that is more or less vector sums of the radii of regular polygons): Draw a nonagon (9-gon), label the corners A through I. Then isosceles triangle ABC above is triangle FAB. Construct segment FX along FA with length = length AB = length FG. Angle XFG = angle AFG = 60 degrees, since arc AG = 120. So triangle XFG is equilateral; Segment BX passes through the center, since B is opposite side FG; Angle XBF = 10 degrees; Angle BXF = 180 - 20 - 10 = 150 degrees, quod erat inveniendum. Roger at MIT-DSPG or ...!decvax!genrad!mit-eddie!rlh