Question

homework 1.5: angle pairs
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1. find the missing measure.
2. find the missing measure.
3. find the missing measures.
4. if the measure of an angle is 13°, find the measure of its supplement.
5. if the measure of an angle is 38°, find the measure of its complement.
6. ∠1 and ∠2 form a linear pair. if m∠1=(5x + 9)° and m∠2=(3x + 11)°, find the measure of each angle.
7. ∠1 and ∠2 are vertical angles. if m∠1=(17x + 1)° and m∠2=(20x - 14)°, find m∠2.
8. ∠k and ∠l are complementary angles. if m∠k=(3x + 3)° and m∠l=(10x - 4)°, find the measure of each angle.

if m∠p is three less than twice the measure of ∠q, and ∠p and ∠q are supplementary angles, find each angle measure.
if m∠b is two more than three times the measure of ∠c, and ∠b and ∠c are complementary angles, find each angle measure.

Explanation:

Step1: Recall angle - pair properties

• Complementary angles add up to 90°. Supplementary angles add up to 180°. Vertical angles are equal. Linear - pair angles add up to 180°.

Step2: Solve problem 1

Since the two angles are complementary (assuming they form a right - angle), if one angle is 65°, then \(x = 90 - 65\).
\(x=25^{\circ}\)

Step3: Solve problem 2

Vertical angles are equal. So \(x = 51^{\circ}\)

Step4: Solve problem 3

The two angles are supplementary. So \(x=180 - 107=73^{\circ}\), and since the other angle \(y\) is vertical to \(x\), \(y = 73^{\circ}\)

Step5: Solve problem 4

The supplement of an angle \(\theta\) is \(180-\theta\). If \(\theta = 13^{\circ}\), then the supplement is \(180 - 13=167^{\circ}\)

Step6: Solve problem 5

The complement of an angle \(\theta\) is \(90-\theta\). If \(\theta = 38^{\circ}\), then the complement is \(90 - 38 = 52^{\circ}\)

Step7: Solve problem 6

Since \(\angle1\) and \(\angle2\) form a linear pair, \(m\angle1 + m\angle2=180^{\circ}\). So \((5x + 9)+(3x + 11)=180\).
Combining like - terms: \(8x+20 = 180\).
Subtract 20 from both sides: \(8x=160\).
Divide both sides by 8: \(x = 20\).
\(m\angle1=(5\times20 + 9)^{\circ}=109^{\circ}\), \(m\angle2=(3\times20 + 11)^{\circ}=71^{\circ}\)

Step8: Solve problem 7

Since \(\angle1\) and \(\angle2\) are vertical angles, \(m\angle1=m\angle2\). So \(17x + 1=20x-14\).
Subtract \(17x\) from both sides: \(1 = 3x-14\).
Add 14 to both sides: \(15 = 3x\).
Divide both sides by 3: \(x = 5\).
\(m\angle2=(20\times5-14)^{\circ}=86^{\circ}\)

Step9: Solve problem 8

Since \(\angle K\) and \(\angle L\) are complementary, \(m\angle K + m\angle L=90^{\circ}\). So \((3x + 3)+(10x-4)=90\).
Combining like - terms: \(13x-1 = 90\).
Add 1 to both sides: \(13x=91\).
Divide both sides by 13: \(x = 7\).
\(m\angle K=(3\times7 + 3)^{\circ}=24^{\circ}\), \(m\angle L=(10\times7-4)^{\circ}=66^{\circ}\)

Step10: Solve problem 9

Let \(m\angle Q=x\). Then \(m\angle P = 2x-3\). Since \(\angle P\) and \(\angle Q\) are supplementary, \(m\angle P+m\angle Q = 180^{\circ}\).
So \((2x-3)+x=180\).
Combining like - terms: \(3x-3 = 180\).
Add 3 to both sides: \(3x=183\).
Divide both sides by 3: \(x = 61^{\circ}\).
\(m\angle Q = 61^{\circ}\), \(m\angle P=2\times61-3=119^{\circ}\)

Step11: Solve problem 10

Let \(m\angle C=x\). Then \(m\angle B = 3x + 2\). Since \(\angle B\) and \(\angle C\) are complementary, \(m\angle B+m\angle C=90^{\circ}\).
So \((3x + 2)+x=90\).
Combining like - terms: \(4x+2 = 90\).
Subtract 2 from both sides: \(4x=88\).
Divide both sides by 4: \(x = 22^{\circ}\).
\(m\angle C = 22^{\circ}\), \(m\angle B=3\times22 + 2=68^{\circ}\)

Answer:

1. \(25^{\circ}\)
2. \(51^{\circ}\)
3. \(x = 73^{\circ},y = 73^{\circ}\)
4. \(167^{\circ}\)
5. \(52^{\circ}\)
6. \(m\angle1 = 109^{\circ},m\angle2 = 71^{\circ}\)
7. \(86^{\circ}\)
8. \(m\angle K = 24^{\circ},m\angle L = 66^{\circ}\)
9. \(m\angle Q = 61^{\circ},m\angle P = 119^{\circ}\)
10. \(m\angle C = 22^{\circ},m\angle B = 68^{\circ}\)