Question

find the angle measure.
a. ∠1 is a complement of ∠2 and m∠1 = 49°. find m∠2
b. ∠3 is a supplement of ∠4 and m∠3 = 119°. find m∠4
c. ∠5 and ∠6 are vertical angles and m∠5 = 33°. find m∠6
d. ∠7 and ∠8 are linear angles and m∠7 is 4 times that of m∠8. find m∠8.
e. ∠1 is a supplement of ∠2 and m∠2 = 31°. find m∠1
f. ∠3 is a complement of ∠4 and m∠3 = 12°. find m∠4
g. find m∠abd.
h. find m∠dbc.
i. find m∠xyw.
j. find m∠wyz.

Explanation:

Step1: Recall complementary - angle property

Complementary angles sum to 90°. Given \(m\angle1 = 49^{\circ}\), then \(m\angle2=90^{\circ}-m\angle1\).
\(m\angle2 = 90 - 49=41^{\circ}\)

Step2: Recall supplementary - angle property

Supplementary angles sum to 180°. Given \(m\angle3 = 119^{\circ}\), then \(m\angle4=180^{\circ}-m\angle3\).
\(m\angle4 = 180 - 119 = 61^{\circ}\)

Step3: Recall vertical - angle property

Vertical angles are equal. Given \(m\angle5 = 33^{\circ}\), then \(m\angle6=m\angle5 = 33^{\circ}\)

Step4: Let \(m\angle8=x\), then \(m\angle7 = 4x\)

Since \(\angle7\) and \(\angle8\) are linear angles, \(m\angle7 + m\angle8=180^{\circ}\). So \(4x+x=180^{\circ}\), \(5x = 180^{\circ}\), \(x=\frac{180}{5}=36^{\circ}\), so \(m\angle8 = 36^{\circ}\)

Step5: Recall supplementary - angle property

Given \(m\angle2 = 31^{\circ}\), then \(m\angle1=180^{\circ}-m\angle2\).
\(m\angle1 = 180 - 31=149^{\circ}\)

Step6: Recall complementary - angle property

Given \(m\angle3 = 12^{\circ}\), then \(m\angle4=90^{\circ}-m\angle3\).
\(m\angle4 = 90 - 12 = 78^{\circ}\)

Step7: Since \(\angle ABD\) and \(\angle DBC\) are linear angles, \((\frac{1}{2}x + 20)+(2x - 10)=180\)

Combine like - terms: \(\frac{1}{2}x+2x+20 - 10 = 180\), \(\frac{1}{2}x+2x+10 = 180\), \(\frac{1 + 4}{2}x=180 - 10\), \(\frac{5}{2}x=170\), \(x=\frac{170\times2}{5}=68\). Then \(m\angle ABD=\frac{1}{2}\times68 + 20=34 + 20 = 54^{\circ}\)

Step8: \(m\angle DBC=2\times68-10=136 - 10 = 126^{\circ}\)

Step9: Since \(\angle XYW\) and \(\angle WYZ\) are complementary, \((1\frac{1}{4}x - 10)+(\frac{3}{4}x + 20)=90\)

Combine like - terms: \((\frac{5}{4}x+\frac{3}{4}x)+(-10 + 20)=90\), \(2x+10 = 90\), \(2x=90 - 10 = 80\), \(x = 40\). Then \(m\angle XYW=1\frac{1}{4}\times40-10=\frac{5}{4}\times40-10=50 - 10 = 40^{\circ}\)

Step10: \(m\angle WYZ=\frac{3}{4}\times40 + 20=30 + 20 = 50^{\circ}\)

Answer:

A. \(41^{\circ}\)
B. \(61^{\circ}\)
C. \(33^{\circ}\)
D. \(36^{\circ}\)
E. \(149^{\circ}\)
F. \(78^{\circ}\)
G. \(54^{\circ}\)
H. \(126^{\circ}\)
I. \(40^{\circ}\)
J. \(50^{\circ}\)