Question

find the value of x.
6.
in the exercises below, determine whether ∠f and ∠g are complementary, supplementary or neither.

1. m∠f = 29°, m∠g = 61°
2. m∠f = 159°, m∠g = 11°
3. m∠f = 37°, m∠g = 37°
4. the measure of ∠m is 56°. find the measure of a complement and a supplement of ∠m.
5. comp -

supp -
use the picture below to answer 11 - 13.

1. find the m∠1.
2. find the m∠2.
3. find the value of y.

Explanation:

Step1: Solve for x in problem 6

Assume the sum of the angles at the vertex is 90° (if they are complementary). So, \(25+(4x - 3)=90\). First, simplify the left - hand side: \(25+4x-3 = 4x + 22\). Then we have the equation \(4x+22 = 90\). Subtract 22 from both sides: \(4x=90 - 22=68\). Divide both sides by 4: \(x = 17\).

Step2: Determine angle relationships in problems 7 - 9

Problem 7

Complementary angles add up to 90°. Since \(m\angle F+m\angle G=29^{\circ}+61^{\circ}=90^{\circ}\), \(\angle F\) and \(\angle G\) are complementary.

Problem 8

Supplementary angles add up to 180°. Since \(m\angle F+m\angle G=159^{\circ}+11^{\circ}=170^{\circ}
eq180^{\circ}\) and \(m\angle F+m\angle G
eq90^{\circ}\), \(\angle F\) and \(\angle G\) are neither complementary nor supplementary.

Problem 9

Since \(m\angle F+m\angle G=37^{\circ}+37^{\circ}=74^{\circ}
eq90^{\circ}\) and \(m\angle F+m\angle G
eq180^{\circ}\), \(\angle F\) and \(\angle G\) are neither complementary nor supplementary.

Step3: Find complement and supplement of \(\angle M\) in problem 10

The complement of an angle \(\theta\) is \(90^{\circ}-\theta\) and the supplement is \(180^{\circ}-\theta\). For \(\angle M = 56^{\circ}\), the complement is \(90^{\circ}-56^{\circ}=34^{\circ}\) and the supplement is \(180^{\circ}-56^{\circ}=124^{\circ}\).

Step4: Solve for angles and y in problems 11 - 13

Problem 11

\(\angle1\) and the \(110^{\circ}\) angle are vertical angles. Vertical angles are equal. So \(m\angle1 = 110^{\circ}\).

Problem 12

\(\angle2\) and the \(110^{\circ}\) angle are supplementary. So \(m\angle2=180^{\circ}-110^{\circ}=70^{\circ}\).

Problem 13

\(\angle2=(3y + 10)^{\circ}\) and \(m\angle2 = 70^{\circ}\). So we set up the equation \(3y+10=70\). Subtract 10 from both sides: \(3y=70 - 10 = 60\). Divide both sides by 3: \(y = 20\).

Answer:

1. \(x = 17\)
2. Complementary
3. Neither
4. Neither
5. Comp - \(34^{\circ}\), Supp - \(124^{\circ}\)
6. \(110^{\circ}\)
7. \(70^{\circ}\)
8. \(y = 20\)