Question 1023440: Point X is on side Line AC of Triangle ABC such that angle AXB =angle ABX, and angle ABC - angle ACB = 39. Find angle XBC in degrees.
Answer by simranb(63)   (Show Source):
You can put this solution on YOUR website! Given angle AXB =angle ABX and angle ABC - angle ACB = 39
Let angle AXB and angle ABX be 'a degrees'.
Consider the triangle AXB,
angle AXB + angle ABX + angle BAX = 180 ( sum of all three angles of a triangle is 180)
a + a + angle BAX = 180
angle BAX = 180-2a
Let angle ABC be 'x' and angle BCA be 'y'.
Consider triangle ABC
angle ABC + angle BCA + angle BAC = 180
x + y + (180-2a) = 180 ( angle BAC is same as the angle BAX)
x+y-2a=0
x+y = 2a ------------ Eq 1
It is given that, angle ABC - angle ACB = 39
so, x-y = 39 --------------Eq 2
Adding Equations 1 and 2 gives,
2x = 39+2a
x = (39+2a)/2
You must note that angle XBC is angle ABC- angle ABX, that is nothing but
angle XBC = x-a
= (39+2a)/2 - a
= (39+2a-2a)/2
=39/2
= 19.5 degrees
Hope this helps!
Cheers!
Simran Bhuria
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