Question

3
m∠acb =
m∠abc =

Explanation:

Step1: Find the value of \( x \)

We know that the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
For \(\triangle ABC\), \(\angle CAD = 103^{\circ}\), \(\angle ACB=(3x - 14)^{\circ}\), and \(\angle ABC=(3x + 9)^{\circ}\)
By the exterior angle theorem: \(\angle CAD=\angle ACB+\angle ABC\)
Substitute the expressions: \(103=(3x - 14)+(3x + 9)\)
Simplify the right - hand side: \(103 = 3x-14 + 3x+9\)
Combine like terms: \(103=6x-5\)
Add \(5\) to both sides: \(103 + 5=6x-5 + 5\), so \(108 = 6x\)
Divide both sides by \(6\): \(x=\frac{108}{6}=18\)

Step2: Calculate \(\angle ACB\)

Substitute \(x = 18\) into the formula for \(\angle ACB\):
\(\angle ACB=(3x - 14)^{\circ}\)
\(\angle ACB=(3\times18-14)^{\circ}\)
First, calculate \(3\times18 = 54\), then \(54-14=40\)
So \(\angle ACB = 40^{\circ}\)

Step3: Calculate \(\angle ABC\)

Substitute \(x = 18\) into the formula for \(\angle ABC\):
\(\angle ABC=(3x + 9)^{\circ}\)
\(\angle ABC=(3\times18+9)^{\circ}\)
First, calculate \(3\times18 = 54\), then \(54 + 9=63\)
So \(\angle ABC = 63^{\circ}\)

Answer:

\(m\angle ACB = 40^{\circ}\), \(m\angle ABC=63^{\circ}\)