[net.math] 80-80-20 Isosceles Triangle Problem

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