Question

4. In the diagram below, ABC is shown with AC extended through point D. If mzBCD = 6x + 2,

m/BAC = 3x + 15, and mLABC = 2x - 1, what is the value of x?

Answer 1

x = 12

Explanation:

Exterior Angle Theorem

The interior angles of a triangle sum to 180°. Angles on a straight line sum to 180°. Therefore, the exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles of the triangle.

Given angles:

• 
  `\textsf{Exterior angle}: \quad m \angle BCD = 6x+2`
• 
  `\textsf{Non-adjacent interior angle}: \quad m \angle BAC= 3x+15`
• 
  `\textsf{Non-adjacent interior angle}: \quad m \angle ABC= 2x-1`

Apply the exterior angle theorem and solve for x:

`\implies m \angle BCD=m \angle BAC+m \angle ABC`

`\implies 6x+2=3x+15+2x-1`

`\implies 6x+2=3x+2x+15-1`

`\implies 6x+2=5x+14`

`\implies 6x+2-5x=5x+14-5x`

`\implies x+2=14`

`\implies x+2-2=14-2`

`\implies x=12`

Answer 2

• x = 12

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According to the diagram, angle BCD is exterior angle and angles BAC & ABC are remote interior angles of the triangle ABC.

As we know, the exterior angle is the sum of remote interior angles.

Set this as equation and solve for x

• m∠BCD = m∠BAC + m∠ABC

• 6x + 2 = 3x + 15 + 2x - 1
• 6x + 2 = 5x + 14
• 6x - 5x = 14 - 2
• x = 12