Question

for questions #4 - 6, refer to the diagram at the right, where ah bisects ∠mat. m∠mah = 3(x - 4) and m∠hat=\frac{1}{2}(4x + 6). 4. find x. 5. find m∠hat. 6. find m∠mat.

Explanation:

Step1: Recall angle - bisector property

Since \(AH\) bisects \(\angle MAT\), then \(m\angle MAH=m\angle HAT\).
So, \(3(x - 4)=\frac{1}{2}(4x + 6)\).

Step2: Expand the left - hand side

Expand \(3(x - 4)\) to get \(3x-12\). The equation becomes \(3x-12=\frac{1}{2}(4x + 6)\).

Step3: Expand the right - hand side

Expand \(\frac{1}{2}(4x + 6)\) to get \(2x+3\). So, \(3x-12 = 2x+3\).

Step4: Solve for \(x\)

Subtract \(2x\) from both sides: \(3x-2x-12=2x - 2x+3\), which simplifies to \(x-12 = 3\). Then add 12 to both sides: \(x=3 + 12\), so \(x = 15\).

Step5: Find \(m\angle HAT\)

Substitute \(x = 15\) into the expression for \(m\angle HAT\), \(m\angle HAT=\frac{1}{2}(4x + 6)\). First, calculate \(4x+6\) when \(x = 15\): \(4\times15+6=60 + 6=66\). Then \(m\angle HAT=\frac{1}{2}\times66 = 33^{\circ}\).

Step6: Find \(m\angle MAT\)

Since \(m\angle MAT=m\angle MAH+m\angle HAT\) and \(m\angle MAH=m\angle HAT\), \(m\angle MAT = 2m\angle HAT\). So \(m\angle MAT=2\times33^{\circ}=66^{\circ}\).

Answer:

1. \(x = 15\)
2. \(m\angle HAT=33^{\circ}\)
3. \(m\angle MAT=66^{\circ}\)