Mathematics Question and Solution
Notice, triangle ABC is isosceles.
Let BC, the base of triangle ABC = 1 unit.
a = ½(180-45)
a = ½(135)
a = 67.5°
a is angle ABC = angle ACB.
b = a-45
b = 22.5°
b is angle DBE.
c = a-30
c = 37.5°
c is angle BCD.
d = 180-45-67.5
d = 67.5°
d is angle BEC.
(1/sin67.5) = (e/sin45)
e = 0.7653668647 units.
e is CE.
f = 180-67.5-37.5
f = 180-105
f = 75°
f is angle BDC.
(1/sin75) = (g/sin67.5)
g = 0.9564704736 units.
g is CD.
h² = 0.9564704736²+0.7653668647²-2*0.9564704736*0.7653668647cos30
h = 0.4823619098 units.
h is DE.
(0.4823619098/sin30) = (0.9564704736/sinj)
j = 82.5°
k = 180-j
k = 97.5°
k is angle CED.
Therefore, the required angle, ? is;
? = k-d
? = 97.5-67.5
? = 30°
