Find x: m<ABC=(6x+18) , m<BAC(8x-3) , m<BCD=(15x+6)

Answer 1

If we take a triangle ABD and draw a segment BC from vertex B to some point on side AD we have figure with two triangles sharing the common side BC. [based on the comment made by James S.]

In this case ∠BCD is an exterior angle to triangle ABC, which means that m∠BCD = m∠ABC + m∠BAC.

We can use this to solve for x as follows:

m∠BCD = m∠ABC + m∠BAC means that

15x + 6 = (6x + 18) + (8x – 3) = 14x +15

Subtracting 14x from both sides gives

15x – 14x + 6 =  x + 6 = 15

Subtracting 6 from both sides gives

x = 15 – 6

answer

x = 9

check answer

15(9) + 6 = (6(9) + 18) + (8(9) – 3)

135 +6 = (54+ 18) + (72 – 3)

141 = 72 + 69