Point D is on side AC of triangel ABC such that angle ADB = angle ABD and angle ABC - angle ACB = 37 degrees. Find angle DBC in degrees.
+129845 A
D
B C
Angle ACB = x
Angle ABC = x + 37
Angle A = 180 - (2x + 37) = 143 - 2x
And since angle ABD = angle ADB then the measure of each of thee angles = (143 -2x) / 2
So angle BDC = 180 - (143-2x)/2
And angle DBC = angle ABC - angle ABD = (x + 37) - (143 -2x) / 2
So in triangle BDC
ACB + BDC + DBC =180
x + 180 - (143 -2x) / 2 + (x + 37) - (143 -2x) / 2 = 180
Simplifying this we get
4x + 74 = 180
4x = 106
x = 106/4 = 53 / 2 = 26.5°
So angle DBC = (26.5 + 37) - (143 - 2(26.5) ) / 2 = 18.5°
