Question

if m∠cdf=(3x + 14)°, m∠fde=(5x - 2)°, and m∠cde=(10x - 18)°, find each measure. m∠cdf = m∠fde = m∠cde = x =

Explanation:

Step1: Set up equation based on angle - addition

Since \(\angle CDE=\angle CDF + \angle FDE\), we have \((3x + 14)^{\circ}+(5x-2)^{\circ}=(10x - 18)^{\circ}\).

Step2: Combine like - terms on the left - hand side

\((3x+5x)+(14 - 2)=8x + 12\). So the equation becomes \(8x+12 = 10x-18\).

Step3: Solve for \(x\)

Subtract \(8x\) from both sides: \(12=10x - 8x-18\), which simplifies to \(12 = 2x-18\). Then add 18 to both sides: \(12 + 18=2x\), so \(30 = 2x\). Divide both sides by 2, we get \(x = 15\).

Step4: Find \(m\angle CDF\)

Substitute \(x = 15\) into the expression for \(m\angle CDF\): \(m\angle CDF=3x + 14=3\times15+14=45 + 14=59^{\circ}\).

Step5: Find \(m\angle FDE\)

Substitute \(x = 15\) into the expression for \(m\angle FDE\): \(m\angle FDE=5x-2=5\times15-2=75 - 2=73^{\circ}\).

Step6: Find \(m\angle CDE\)

Substitute \(x = 15\) into the expression for \(m\angle CDE\): \(m\angle CDE=10x-18=10\times15-18=150 - 18=132^{\circ}\).

Answer:

\(x = 15\), \(m\angle CDF=59^{\circ}\), \(m\angle FDE=73^{\circ}\), \(m\angle CDE=132^{\circ}\)