Question

1.3 homework
name:
date:
per:

1. use the diagram below to complete each part.

a) name the vertex of ∠1.
b) name the sides of ∠2.
c) another name for ∠5.
d) classify each angle:
∠fbc, ∠ebf, ∠abc
e) name an angle bisector.
f) if m∠ebd = 36° and m∠dbc = 108°, find m∠ebc.
g) if m∠ebf = 117°, find m∠abe.
h) if bf ⊥ ac

1. if m∠mkl = 83°, m∠jkl = 127°, and m∠jkm=(9x - 10)°, find the value of x.
2. if m∠efh=(5x + 17)°, m∠hfg = 42°, and m∠efg=(18x + 11)°, find each measure.
3. if m∠rst=(12x - 1)°, m∠rsu=(9x - 15)°, and m∠ust = 53°, find each measure.
4. if m∠wxz=(5x + 3)°, m∠zxy=(8x - 4)°, and ∠wxy is a right - angle, find each measure.

Explanation:

Step1: Analyze angle - related equations

For problem 2:
We know that \(\angle MKZ\) and \(\angle KLM\) are related to \(\angle KJM\). If we assume there is a linear - pair or some angle - addition relationship. Let's assume \(\angle MKZ+\angle KLM+\angle KJM = 360^{\circ}\) (if they are around a point) or some other geometric relationship based on the figure. Given \(\angle MKZ = 83^{\circ}\), \(\angle KLM=127^{\circ}\), and \(\angle KJM=(9x - 10)^{\circ}\). Then \(83 + 127+(9x - 10)=360\) (assuming they are angles around a point).
First, simplify the left - hand side: \(83+127 - 10+9x=360\), \(200 - 10+9x=360\), \(190+9x=360\).

Step2: Solve for \(x\)

Subtract 190 from both sides of the equation \(190+9x=360\): \(9x=360 - 190\), \(9x = 170\), \(x=\frac{170}{9}\approx18.89\).

For problem 3:
We know that \(\angle EFH=(5x + 17)^{\circ}\), \(\angle HFG = 42^{\circ}\), and \(\angle EFG=(18x+11)^{\circ}\). Since \(\angle EFG=\angle EFH+\angle HFG\) (angle - addition postulate), we have the equation \((5x + 17)+42=(18x+11)\).
Simplify the left - hand side: \(5x+59 = 18x+11\).
Subtract \(5x\) from both sides: \(59=18x - 5x+11\), \(59 = 13x+11\).
Subtract 11 from both sides: \(13x=59 - 11\), \(13x = 48\), \(x=\frac{48}{13}\approx3.69\).
Then \(\angle EFH=5x + 17=5\times\frac{48}{13}+17=\frac{240}{13}+17=\frac{240 + 221}{13}=\frac{461}{13}\approx35.46^{\circ}\), \(\angle EFG=18x + 11=18\times\frac{48}{13}+11=\frac{864}{13}+11=\frac{864+143}{13}=\frac{1007}{13}\approx77.46^{\circ}\).

For problem 4:
We know that \(\angle RST=(12x - 1)^{\circ}\), \(\angle RSU=(9x - 15)^{\circ}\), and \(\angle UST = 53^{\circ}\). Since \(\angle RST=\angle RSU+\angle UST\) (angle - addition postulate), we have the equation \(12x-1=(9x - 15)+53\).
Simplify the right - hand side: \(12x-1=9x+38\).
Subtract \(9x\) from both sides: \(12x-9x-1=38\), \(3x-1 = 38\).
Add 1 to both sides: \(3x=39\), \(x = 13\).
Then \(\angle RST=12x-1=12\times13-1=156 - 1=155^{\circ}\), \(\angle RSU=9x - 15=9\times13-15=117 - 15 = 102^{\circ}\).

For problem 5:
Since \(\angle WXY\) is a right - angle (\(90^{\circ}\)), and \(\angle WXZ=(5x + 3)^{\circ}\), \(\angle ZXY=(8x - 4)^{\circ}\), and \(\angle WXY=\angle WXZ+\angle ZXY\) (angle - addition postulate). Then \((5x + 3)+(8x - 4)=90\).
Simplify the left - hand side: \(5x+8x+3 - 4=90\), \(13x-1=90\).
Add 1 to both sides: \(13x=91\), \(x = 7\).
Then \(\angle WXZ=5x + 3=5\times7+3=35 + 3=38^{\circ}\), \(\angle ZXY=8x - 4=8\times7-4=56 - 4=52^{\circ}\).

Answer:

Problem 2: \(x=\frac{170}{9}\)
Problem 3: \(x=\frac{48}{13}\), \(\angle EFH=\frac{461}{13}^{\circ}\), \(\angle EFG=\frac{1007}{13}^{\circ}\)
Problem 4: \(x = 13\), \(\angle RST=155^{\circ}\), \(\angle RSU=102^{\circ}\)
Problem 5: \(x = 7\), \(\angle WXZ=38^{\circ}\), \(\angle ZXY=52^{\circ}\)