Question

1. in the diagram below, ab and cd are parallel and m∠6=(3x + 3)° and m∠3=(7x + 27)°. determine m∠5.

Explanation:

Step1: Use the property of parallel lines

Since \(AB\parallel CD\), \(\angle3\) and \(\angle6\) are corresponding angles. So \(3x + 3=7x+27\).

Step2: Solve the equation for \(x\)

Subtract \(3x\) from both sides: \(3 = 4x+27\).
Subtract \(27\) from both sides: \(4x=3 - 27=- 24\).
Divide both sides by \(4\): \(x=-6\).

Step3: Find \(m\angle6\)

Substitute \(x = - 6\) into \(m\angle6=(3x + 3)^{\circ}\), \(m\angle6=3\times(-6)+3=-18 + 3=-15^{\circ}\) (This is wrong, actually we should use the property that \(\angle5\) and \(\angle6\) are supplementary (\(\angle5+\angle6 = 180^{\circ}\)) since they are adjacent - angles on a straight line. First, correct the equation from step 1: \(\angle3\) and \(\angle6\) are same - side interior angles, so \(m\angle3+m\angle6 = 180^{\circ}\). Then \(3x + 3+7x+27=180\). Combine like terms: \(10x+30 = 180\). Subtract \(30\) from both sides: \(10x=150\). Divide by \(10\): \(x = 15\).

Step4: Find \(m\angle6\)

Substitute \(x = 15\) into \(m\angle6=(3x + 3)^{\circ}\), \(m\angle6=3\times15+3=45 + 3=48^{\circ}\).

Step5: Find \(m\angle5\)

Since \(\angle5+\angle6 = 180^{\circ}\) (linear - pair postulate), \(m\angle5=180^{\circ}-m\angle6\). Substitute \(m\angle6 = 48^{\circ}\), \(m\angle5=180 - 48=132^{\circ}\).

Answer:

\(132^{\circ}\)