Question

angle bisectors

1. if $overrightarrow{bx}$ bisects $angle abc$, $mangle abx = 5x$, and $angle xbc=3x + 10$, find $x$, $mangle abx$, and $mangle abc.
2. if $overrightarrow{kn}$ bisects $angle jkl$, $mangle jkn = 4x - 16$, and $mangle nkl = 2x+6$, find $x$ and $mangle jkl.
3. if $overrightarrow{ns}$ bisects $angle mno$, $mangle mns = 24$, and $mangle mno = 4x - 12$, find $x$ and $mangle mno.

vertical angles

1. if $mangle 1=x + 10$ and $mangle 2 = 4x-35$, find $x$ and $mangle 1.
2. $angle 3$ and $angle 4$ are vertical angles. $mangle 3 = 3x + 8$ and $mangle 4 = 5x-20$, find $x$ and $mangle 4.

Explanation:

Step1: Use angle - bisector property for problem 6

Since $\overrightarrow{BX}$ bisects $\angle ABC$, then $m\angle ABX=m\angle XBC$. So, $5x = 3x+10$.
Subtract $3x$ from both sides: $5x - 3x=3x + 10-3x$, which gives $2x=10$.
Divide both sides by 2: $x = 5$.
$m\angle ABX=5x=5\times5 = 25$.
$m\angle ABC=m\angle ABX + m\angle XBC=25+25 = 50$.

Step2: Use angle - bisector property for problem 7

Since $\overrightarrow{KN}$ bisects $\angle JKL$, then $m\angle JKN=m\angle NKL$. So, $4x-16 = 2x + 6$.
Subtract $2x$ from both sides: $4x-2x-16=2x-2x + 6$, which gives $2x-16 = 6$.
Add 16 to both sides: $2x-16 + 16=6 + 16$, so $2x=22$.
Divide both sides by 2: $x = 11$.
$m\angle JKL=m\angle JKN + m\angle NKL=(4x-16)+(2x + 6)=4\times11-16+2\times11 + 6=44-16+22 + 6=56$.

Step3: Use angle - bisector property for problem 8

Since $\overrightarrow{NS}$ bisects $\angle MNO$, then $m\angle MNO = 2m\angle MNS$. Given $m\angle MNS = 24$ and $m\angle MNO=4x-12$, we have $4x-12=2\times24$.
$4x-12 = 48$.
Add 12 to both sides: $4x-12 + 12=48+12$, so $4x=60$.
Divide both sides by 4: $x = 15$.
$m\angle MNO=4x-12=4\times15-12=48$.

Step4: Use vertical - angle property for problem 9

Since $\angle1$ and $\angle2$ are vertical angles, $m\angle1=m\angle2$. So, $x + 10=4x-35$.
Subtract $x$ from both sides: $x-x + 10=4x-x-35$, which gives $10 = 3x-35$.
Add 35 to both sides: $10 + 35=3x-35 + 35$, so $45=3x$.
Divide both sides by 3: $x = 15$.
$m\angle1=x + 10=15+10 = 25$.

Step5: Use vertical - angle property for problem 10

Since $\angle3$ and $\angle4$ are vertical angles, $m\angle3=m\angle4$. So, $3x + 8=5x-20$.
Subtract $3x$ from both sides: $3x-3x + 8=5x-3x-20$, which gives $8 = 2x-20$.
Add 20 to both sides: $8 + 20=2x-20 + 20$, so $28=2x$.
Divide both sides by 2: $x = 14$.
$m\angle4=5x-20=5\times14-20=50$.

Answer:

1. $x = 5$, $m\angle ABX = 25$, $m\angle ABC = 50$
2. $x = 11$, $m\angle JKL = 56$
3. $x = 15$, $m\angle MNO = 48$
4. $x = 15$, $m\angle1 = 25$
5. $x = 14$, $m\angle4 = 50$