practice
example 1
1. find the measures of two supplementary angles if the difference between the measures of the two angles is 35°.
2. ∠e and ∠f are complementary. the measure of ∠e is 54° more than the measure of ∠f. find the measure of each angle.
3. the measure of an angle’s supplement is 76° less than the measure of the angle. find the measures of the angle and its supplement.
4. ∠q and ∠r are complementary. the measure of ∠q is 26° less than the measure of ∠r. find the measure of each angle.
5. the measure of the supplement of an angle is three times the measure of the angle. find the measures of the angle and its supplement.
6. the bascule bridge shown is opening from its horizontal position to its fully vertical position. so far, the bridge has lifted 35° in 21 seconds. at this rate, how much longer will it take for the bridge to reach its vertical position?
example 2
7. rays ba and bc are perpendicular. point d lies in the interior of ∠abc. if m∠abd=(3r + 5)° and m∠dbc=(5r - 27)°, find m∠abd and m∠dbc.
8. wx and yz intersect at point v. if m∠wvy=(4a + 58)° and m∠xvy=(2b - 18)°, find the values of a and b such that wx is perpendicular to yz.
9. refer to the figure at the right. if m∠2=(a + 15)° and m∠3=(a + 35)°, find the value of a such that hl⊥hj.
10. rays da and dc are perpendicular. point b lies in the interior of ∠adc. if m∠adb=(3a + 10)° and m∠bdc = 13a°, find a, m∠adb, and m∠bdc.

Question

Accepted Answer

1.
# Explanation:
## Step1: Let the two supplementary angles be $x$ and $y$ ($x>y$). Supplementary - angle property
We know that $x + y=180^{\circ}$ and $x - y = 35^{\circ}$.
## Step2: Add the two equations
$(x + y)+(x - y)=180^{\circ}+35^{\circ}$, which simplifies to $2x=215^{\circ}$, so $x = 107.5^{\circ}$.
## Step3: Find $y$
Substitute $x = 107.5^{\circ}$ into $x + y=180^{\circ}$, we get $y=180^{\circ}-107.5^{\circ}=72.5^{\circ}$.
# Answer:
The two angles are $107.5^{\circ}$ and $72.5^{\circ}$

2.
# Explanation:
## Step1: Let $m\angle F=x$ and $m\angle E = y$. Complementary - angle property
Since $\angle E$ and $\angle F$ are complementary, $x + y=90^{\circ}$, and $y=x + 54^{\circ}$.
## Step2: Substitute $y$ into the first equation
$x+(x + 54^{\circ})=90^{\circ}$, which simplifies to $2x=90^{\circ}-54^{\circ}=36^{\circ}$, so $x = 18^{\circ}$.
## Step3: Find $y$
Substitute $x = 18^{\circ}$ into $y=x + 54^{\circ}$, we get $y=18^{\circ}+54^{\circ}=72^{\circ}$.
# Answer:
$m\angle F = 18^{\circ}$, $m\angle E=72^{\circ}$

3.
# Explanation:
## Step1: Let the angle be $x$ and its supplement be $y$. Supplementary - angle property
We have $x + y=180^{\circ}$ and $y=x - 76^{\circ}$.
## Step2: Substitute $y$ into the first equation
$x+(x - 76^{\circ})=180^{\circ}$, which simplifies to $2x=180^{\circ}+76^{\circ}=256^{\circ}$, so $x = 128^{\circ}$.
## Step3: Find $y$
Substitute $x = 128^{\circ}$ into $y=x - 76^{\circ}$, we get $y=128^{\circ}-76^{\circ}=52^{\circ}$.
# Answer:
The angle is $128^{\circ}$, its supplement is $52^{\circ}$

4.
# Explanation:
## Step1: Let $m\angle R=x$ and $m\angle Q = y$. Complementary - angle property
Since $\angle Q$ and $\angle R$ are complementary, $x + y=90^{\circ}$, and $y=x - 26^{\circ}$.
## Step2: Substitute $y$ into the first equation
$x+(x - 26^{\circ})=90^{\circ}$, which simplifies to $2x=90^{\circ}+26^{\circ}=116^{\circ}$, so $x = 58^{\circ}$.
## Step3: Find $y$
Substitute $x = 58^{\circ}$ into $y=x - 26^{\circ}$, we get $y=58^{\circ}-26^{\circ}=32^{\circ}$.
# Answer:
$m\angle R = 58^{\circ}$, $m\angle Q=32^{\circ}$

5.
# Explanation:
## Step1: Let the angle be $x$ and its supplement be $y$. Supplementary - angle property
We have $x + y=180^{\circ}$ and $y = 3x$.
## Step2: Substitute $y$ into the first equation
$x+3x=180^{\circ}$, which simplifies to $4x=180^{\circ}$, so $x = 45^{\circ}$.
## Step3: Find $y$
Substitute $x = 45^{\circ}$ into $y = 3x$, we get $y=3\times45^{\circ}=135^{\circ}$.
# Answer:
The angle is $45^{\circ}$, its supplement is $135^{\circ}$

6.
# Explanation:
## Step1: Calculate the remaining angle
The bridge needs to move from $35^{\circ}$ to $90^{\circ}$, so the remaining angle $\theta=90^{\circ}-35^{\circ}=55^{\circ}$.
## Step2: Find the rate of movement
The rate of movement $r=\frac{35^{\circ}}{21\text{ s}}=\frac{5}{3}\text{ degrees per second}$.
## Step3: Calculate the remaining time
Using the formula $t=\frac{\theta}{r}$, we have $t=\frac{55^{\circ}}{\frac{5}{3}\text{ degrees per second}}=33\text{ s}$.
# Answer:
33 seconds

7.
# Explanation:
## Step1: Use the right - angle property
Since $\angle ABC = 90^{\circ}$ and $\angle ABC=\angle ABD+\angle DBC$, we have $(3r + 5)+(5r-27)=90$.
## Step2: Combine like terms
$3r+5r+5 - 27=90$, which simplifies to $8r-22 = 90$.
## Step3: Solve for $r$
Add 22 to both sides: $8r=90 + 22=112$, then $r = 14$.
## Step4: Find $m\angle ABD$ and $m\angle DBC$
$m\angle ABD=3r + 5=3\times14+5=47^{\circ}$, $m\angle DBC=5r-27=5\times14-27=43^{\circ}$.
# Answer:
$m\angle ABD = 47^{\circ}$, $m\angle DBC=43^{\circ}$

8.
# Explanation:
## Step1: Use the perpendicular - angle property
If $\overrightarrow{WX}\perp\overrightarrow{YZ}$, then $m\angle WVY=m\angle XVY = 90^{\circ}$.
For $m\angle WVY=(4a + 58)^{\circ}=90^{\circ}$, solve for $a$: $4a=90 - 58=32$, so $a = 8$.
For $m\angle XVY=(2b-18)^{\circ}=90^{\circ}$, solve for $b$: $2b=90 + 18=108$, so $b = 54$.
# Answer:
$a = 8$, $b = 54$

9.
# Explanation:
## Step1: Use the perpendicular - angle property
If $\overrightarrow{HL}\perp\overrightarrow{HJ}$, then $m\angle 2+m\angle 3 = 90^{\circ}$.
So $(a + 15)+(a + 35)=90$.
## Step2: Combine like terms
$a+a+15 + 35=90$, which simplifies to $2a+50 = 90$.
## Step3: Solve for $a$
Subtract 50 from both sides: $2a=90 - 50=40$, then $a = 20$.
# Answer:
$a = 20$

10.
# Explanation:
## Step1: Use the right - angle property
Since $\angle ADC = 90^{\circ}$ and $\angle ADC=\angle ADB+\angle BDC$, we have $(3a + 10)+13a=90$.
## Step2: Combine like terms
$3a+13a+10=90$, which simplifies to $16a+10 = 90$.
## Step3: Solve for $a$
Subtract 10 from both sides: $16a=90 - 10=80$, then $a = 5$.
## Step4: Find $m\angle ADB$ and $m\angle BDC$
$m\angle ADB=3a + 10=3\times5+10=25^{\circ}$, $m\angle BDC=13a=13\times5=65^{\circ}$.
# Answer:
$a = 5$, $m\angle ADB = 25^{\circ}$, $m\angle BDC=65^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Let the two supplementary angles be \(x\) and \(y\) (\(x>y\)). Supplementary - angle property

Step2: Add the two equations

Step3: Find \(y\)

Step1: Let \(m\angle F=x\) and \(m\angle E = y\). Complementary - angle property

Step2: Substitute \(y\) into the first equation

Step3: Find \(y\)

Step1: Let the angle be \(x\) and its supplement be \(y\). Supplementary - angle property

Step2: Substitute \(y\) into the first equation

Step3: Find \(y\)

Answer: