Question

1. if $overline{cd}perpoverline{ef}$, $mangle ech=(x + 5)^{circ}$ and $mangle hcd=(3x - 7)^{circ}$, find each missing value.
2. $angle1$ and $angle2$ are vertical angles. if $mangle1=(6x + 11)^{circ}$ and $mangle2=(10x - 9)^{circ}$, find $mangle1$.
3. if $overline{bd}$ bisects $angle abc$, $mangle dbc = 79^{circ}$, and $mangle abc=(9x - 4)^{circ}$, find the value of $x$.
4. the measure of $angle g$ is six more than twice the measure of $angle h$. if $angle g$ and $angle h$ are complementary angles, find $mangle h$.

Explanation:

Step1: Solve for \(x\) in problem 17

Since \(\overline{CD}\perp\overline{EF}\), \(\angle ECH+\angle HCD = 90^{\circ}\). So, \((x + 5)+(3x-7)=90\).
Combining like - terms gives \(4x-2 = 90\).
Adding 2 to both sides: \(4x=92\).
Dividing both sides by 4: \(x = 23\).

Step2: Find \(m\angle ECH\) in problem 17

Substitute \(x = 23\) into \(m\angle ECH=(x + 5)^{\circ}\).
\(m\angle ECH=(23 + 5)^{\circ}=28^{\circ}\).

Step3: Find \(m\angle HCD\) in problem 17

Substitute \(x = 23\) into \(m\angle HCD=(3x - 7)^{\circ}\).
\(m\angle HCD=(3\times23-7)^{\circ}=(69 - 7)^{\circ}=62^{\circ}\).

Step4: Find \(m\angle GCF\) in problem 17

\(\angle GCF\) and \(\angle ECH\) are vertical angles. Vertical angles are equal. So \(m\angle GCF=m\angle ECH = 28^{\circ}\).

Step5: Find \(m\angle ECG\) in problem 17

\(\angle ECG = 180^{\circ}-\angle GCF=180 - 28=152^{\circ}\).

Step6: Find \(m\angle GCD\) in problem 17

\(\angle GCD=\angle GCF+\angle FCD\), and \(\angle FCD = 90^{\circ}\), \(\angle GCF = 28^{\circ}\), so \(m\angle GCD=28^{\circ}+90^{\circ}=118^{\circ}\).

Step7: Solve for \(x\) in problem 19

Since \(\overline{BD}\) bisects \(\angle ABC\), then \(m\angle ABC = 2m\angle DBC\).
We know \(m\angle DBC = 79^{\circ}\) and \(m\angle ABC=(9x - 4)^{\circ}\), so \(9x-4=2\times79\).
\(9x-4 = 158\).
Adding 4 to both sides: \(9x=162\).
Dividing both sides by 9: \(x = 18\).

Step8: Solve for \(m\angle H\) in problem 20

Let \(m\angle H=x\). Then \(m\angle G=2x + 6\).
Since \(\angle G\) and \(\angle H\) are complementary, \(m\angle G+m\angle H=90^{\circ}\).
\((2x + 6)+x=90\).
Combining like - terms: \(3x+6 = 90\).
Subtracting 6 from both sides: \(3x=84\).
Dividing both sides by 3: \(x = 28\), so \(m\angle H = 28^{\circ}\).

Answer:

a) \(x = 23\)
b) \(m\angle ECH = 28^{\circ}\)
c) \(m\angle HCD = 62^{\circ}\)
d) \(m\angle GCF = 28^{\circ}\)
e) \(m\angle ECG = 152^{\circ}\)
f) \(m\angle GCD = 118^{\circ}\)

1. \(x = 18\)
2. \(m\angle H = 28^{\circ}\)