Question

1. for each set of angles, determine whether there exists a triangle whose 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 (mangle cnf=(2x + 17)^{circ}) and (mangle rmw=(7x + 28)^{circ}).

a. find the value of x.
b. find (mangle cnf) and (mangle rmn).

Explanation:

Step1: Recall angle - sum property of a triangle

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

Step2: Check option a

For angles 70°, 20°, 90°, we calculate the sum: 70 + 20+90 = 180°. So, a triangle can be formed.

Step3: Check option b

For angles 120°, 50°, 25°, the sum is 120 + 50+25=195°≠180°. So, a triangle cannot be formed.

Step4: Check option c

For angles 57°, 100°, 23°, the sum is 57 + 100+23 = 180°. So, a triangle can be formed.

for question 4a:

Step1: Use the property of corresponding angles

Since KR and TC are parallel and WF is a transversal, ∠CNF and ∠RMW are corresponding angles and are equal. So, 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, - 11 = 5x. Divide both sides by 5: x=- \frac{11}{5}=-2.2

for question 4b:

Step1: Find m∠CNF

Substitute x = - 2.2 into the expression for m∠CNF: m∠CNF=2x + 17=2(-2.2)+17=-4.4 + 17 = 12.6°

Step2: Note the relationship between ∠RMW and ∠RMN

∠RMW and ∠RMN are a linear - pair, so ∠RMN = 180°-∠RMW. Since ∠RMW = ∠CNF = 12.6°, then ∠RMN=180 - 12.6 = 167.4°

Answer:

a. Yes
b. No
c. Yes