For simplicity let all the sides be of length 1 WLOG.
Distance AC can be found via the law of cosines:
d^2 = 1 + 1 - 2cos(136°)
d ~= 1.85436770913357
sin(BCA) ~= sin(136°)/1.85436770913357 ~= .374606593415914
and BCA is acute as ABC is obtuse in triangle ABC
BCA ~= 22.0000000000001°, assume = 22°
Angle ACD is therefore (104 - 22)° = 82°
The distance AD is
(mAD)^2 ~= 1.85436770913357^2 + 1 - 2*1.85436770913357*cos(82°)
~= 3.92252339187661
mAD ~= 1.98053613748313
As AE and ED are each 1:
1.98053613748313^2 = 1 + 1 - 2cos(AED)
AED = arccos((1 + 1 - 1.98053613748313^2) / 2) = 164°
(calculator came up with 163.999999999996)
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Posted by Charlie
on 2018-04-12 10:55:33 |
the trig solution
