Mathematics Question and Solution
Notice.
a = 18+55 = 73°
b = 28+45 = 73°
It implies;
The ascribed triangle is isosceles.
Calculating angle x.
Let the shorter length of the ascribed isosceles triangle be 1 unit.
c = 180-45-55-18
c = 180-118
c = 62°
(1/sin62) = (d/sin45)
d = 0.800847963 units.
cos73 = e/0.800847963
e = 0.2341452842 units.
f = 1-2*e
f = 1-2*0.2341452842
f = 0.5317094316 units.
g = ½(360-73-73)
g = ½(360-146)
g = ½(214)
g = 107°
h = 180-55-73
h = 180-128
h = 52°
(1/sin52) = (j/sin55)
j = 1.0395188651 units.
k = j-d
k = 1.0395188651-0.800847963
k = 0.2386709021 units.
l = 180-g
l = 180-107
l = 73°
m² = 0.2386709021²+0.5317094316²-2*0.2386709021*0.5317094316cos73
m = 0.5152404227 units.
(0.5152404227/sin73) = (0.5317094316/sinn)
n = 80.7057143335°
It implies, required angle x is;
n-h
= 80.7057143335-52
= 28.7057143335°
≈ 28.71°
