Question

review problems

1. for each set of angles, determine whether there exists a triangle with these measures.

a. 70°,20°,90°
b. 120°,50°,25°
c. 57°,100°,23°

1. parallel lines kr and tc and transversal wf are shown, where m∠cnf=(2x + 17)° and m∠rmw=(7x + 28)°.

a. find the value of x.
b. find m∠cnf and m∠rmn.

Explanation:

3.

Step1: Recall angle - sum property of a triangle

The sum of the interior angles of a triangle is always 180°.

Step2: Calculate the sum for part a

For the angles 70°, 20°, 90°, we have \(70 + 20+90=180\). So, a triangle with these angle - measures exists.

Step3: Calculate the sum for part b

For the angles 120°, 50°, 25°, we have \(120 + 50+25 = 195
eq180\). So, a triangle with these angle - measures does not exist.

Step4: Calculate the sum for part c

For the angles 57°, 100°, 23°, we have \(57+100 + 23=180\). So, a triangle with these angle - measures exists.

Step1: Use the property of corresponding angles

Since \(KR\parallel TC\) and \(WF\) is a transversal, \(\angle CNF\) and \(\angle RMW\) are corresponding angles, so \(m\angle CNF=m\angle RMW\). Set up the equation \(2x + 17=7x+28\).

Step2: Solve the equation for \(x\)

Subtract \(2x\) from both sides: \(17 = 5x+28\). Then subtract 28 from both sides: \(17−28=5x\), so \(- 11 = 5x\). Divide both sides by 5, we get \(x=-\frac{11}{5}=-2.2\).

Step3: Find \(m\angle CNF\)

Substitute \(x = - 2.2\) into the expression for \(m\angle CNF\): \(m\angle CNF=2x + 17=2(-2.2)+17=-4.4 + 17 = 12.6^{\circ}\).

Step4: Note the relationship between \(\angle RMW\) and \(\angle RMN\)

\(\angle RMW\) and \(\angle RMN\) are a linear - pair, so \(m\angle RMW+m\angle RMN = 180^{\circ}\). Since \(m\angle RMW=m\angle CNF = 12.6^{\circ}\), then \(m\angle RMN=180 - 12.6=167.4^{\circ}\).

Answer:

a. Yes
b. No
c. Yes

4.