+614 in triangle ABC, a point D is on line AC so that AB=AD and angle ABC- angle ACB=32 degrees. Find angle DBC in degrees.
+129852 Since AB = AD, let angles ADB and ABD = x
And let angle DAB = angle CAB = z
Let angle DBC = y
And let angle ABC = x+ y
Andl let angle ACB = x + y - 32
So...we have the following equations :
(x + y ) + [ (x + y) - 32] + z = 180 ⇒ 2x + 2y + z = 212 ⇒ (2x + z) + 2y = 212 (1)
2x + z = 180 (2)
Sub (2) into (1) and we have that
180 + 2y = 212
2y = 32
y = 16° = angle DBC
