Question

1. if (mangle cdf=(3x + 14)^{circ}), (mangle fde=(5x - 2)^{circ}), and (mangle cde=(10x - 18)^{circ}), find each measure.

(x=)
(mangle cdf=)
(mangle fde=)
(mangle cde=)

1. if (mangle lmp) is 11 degrees more than (mangle nmp) and (mangle nml = 137^{circ}), find each measure.

(mangle lmp=)
(mangle nmp=)

Explanation:

Step1: Set up equation for problem 6

Since \(m\angle CDF + m\angle FDE=m\angle CDE\), we substitute the given angle - expressions: \((3x + 14)+(5x - 2)=10x-18\).

Step2: Simplify the left - hand side of the equation

Combine like terms: \(3x+5x + 14 - 2=8x + 12\). 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 the measure of each angle for problem 6

• \(m\angle CDF=3x + 14=3\times15 + 14=45 + 14 = 59^{\circ}\).
• \(m\angle FDE=5x - 2=5\times15 - 2=75 - 2 = 73^{\circ}\).
• \(m\angle CDE=10x-18=10\times15-18=150 - 18 = 132^{\circ}\).

Step5: Set up equation for problem 7

Let \(m\angle NMP=y\). Then \(m\angle LMP=y + 11\). Since \(m\angle NML=m\angle NMP+m\angle LMP\) and \(m\angle NML = 137^{\circ}\), we have \(y+(y + 11)=137\).

Step6: Simplify the equation for problem 7

Combine like terms: \(2y+11 = 137\).

Step7: Solve for \(y\) (measure of \(\angle NMP\))

Subtract 11 from both sides: \(2y=137 - 11=126\). Divide both sides by 2: \(y = 63^{\circ}\), so \(m\angle NMP = 63^{\circ}\).

Step8: Find the measure of \(\angle LMP\)

\(m\angle LMP=m\angle NMP + 11\), so \(m\angle LMP=63+11 = 74^{\circ}\).

Answer:

For problem 6:
\(x = 15\), \(m\angle CDF = 59^{\circ}\), \(m\angle FDE = 73^{\circ}\), \(m\angle CDE = 132^{\circ}\)
For problem 7:
\(m\angle NMP = 63^{\circ}\), \(m\angle LMP = 74^{\circ}\)