QUESTION IMAGE
Question
use the diagram below to answer questions 1 and 2.
- if (mangle abd = 48^{circ}) and (mangle dbc = 78^{circ}), find (mangle abc).
- if (mangle dbc = 74^{circ}) and (mangle abc = 119^{circ}), find (mangle abd).
- if (mangle pqr = 141^{circ}), find each measure.
- if (mangle def=(7x + 4)^{circ}), (mangle deg=(5x + 1)^{circ}), and (mangle gef = 23^{circ}), find each measure.
- if (mangle jkm = 43^{circ}), (mangle mkl=(8x - 20)^{circ}), and (mangle jkl=(10x - 11)^{circ}), find each measure.
Step1: Use angle - addition postulate for question 1
The angle - addition postulate states that if point \(D\) lies in the interior of \(\angle ABC\), then \(m\angle ABC=m\angle ABD + m\angle DBC\). Given \(m\angle ABD = 48^{\circ}\) and \(m\angle DBC=78^{\circ}\), we have \(m\angle ABC=48^{\circ}+78^{\circ}\).
\(m\angle ABC = 126^{\circ}\)
Step2: Use angle - addition postulate for question 2
Since \(m\angle ABC=m\angle ABD + m\angle DBC\), then \(m\angle ABD=m\angle ABC - m\angle DBC\). Given \(m\angle DBC = 74^{\circ}\) and \(m\angle ABC = 119^{\circ}\), we get \(m\angle ABD=119^{\circ}-74^{\circ}\).
\(m\angle ABD = 45^{\circ}\)
Step3: Use angle - addition postulate for question 3
Since \(m\angle PQR=m\angle PQS + m\angle SQR\), and \(m\angle PQS=(13x + 4)^{\circ}\), \(m\angle SQR=(10x-1)^{\circ}\), \(m\angle PQR = 141^{\circ}\), we have the equation \((13x + 4)+(10x-1)=141\).
First, simplify the left - hand side: \(13x+4 + 10x-1=23x + 3\).
Then solve the equation \(23x+3 = 141\). Subtract 3 from both sides: \(23x=141 - 3=138\). Divide both sides by 23: \(x = 6\).
\(m\angle PQS=13x + 4=13\times6+4=82^{\circ}\)
\(m\angle SQR=10x-1=10\times6-1 = 59^{\circ}\)
Step4: Use angle - addition postulate for question 4
Since \(m\angle DEF=m\angle DEG + m\angle GEF\), we have the equation \((7x + 4)=(5x + 1)+23\).
Simplify the right - hand side: \((5x + 1)+23=5x+24\).
Then solve the equation \(7x + 4=5x+24\). Subtract \(5x\) from both sides: \(7x-5x+4=5x-5x + 24\), \(2x+4 = 24\). Subtract 4 from both sides: \(2x=20\). Divide both sides by 2: \(x = 10\).
\(m\angle DEG=5x + 1=5\times10+1=51^{\circ}\)
\(m\angle DEF=7x + 4=7\times10+4=74^{\circ}\)
Step5: Use angle - addition postulate for question 5
Since \(m\angle JKL=m\angle JKM + m\angle MKL\), we have the equation \((10x-11)=43+(8x - 20)\).
Simplify the right - hand side: \(43+(8x - 20)=8x+23\).
Then solve the equation \(10x-11=8x + 23\). Subtract \(8x\) from both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both b…
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Step1: Use angle - addition postulate for question 1
The angle - addition postulate states that if point \(D\) lies in the interior of \(\angle ABC\), then \(m\angle ABC=m\angle ABD + m\angle DBC\). Given \(m\angle ABD = 48^{\circ}\) and \(m\angle DBC=78^{\circ}\), we have \(m\angle ABC=48^{\circ}+78^{\circ}\).
\(m\angle ABC = 126^{\circ}\)
Step2: Use angle - addition postulate for question 2
Since \(m\angle ABC=m\angle ABD + m\angle DBC\), then \(m\angle ABD=m\angle ABC - m\angle DBC\). Given \(m\angle DBC = 74^{\circ}\) and \(m\angle ABC = 119^{\circ}\), we get \(m\angle ABD=119^{\circ}-74^{\circ}\).
\(m\angle ABD = 45^{\circ}\)
Step3: Use angle - addition postulate for question 3
Since \(m\angle PQR=m\angle PQS + m\angle SQR\), and \(m\angle PQS=(13x + 4)^{\circ}\), \(m\angle SQR=(10x-1)^{\circ}\), \(m\angle PQR = 141^{\circ}\), we have the equation \((13x + 4)+(10x-1)=141\).
First, simplify the left - hand side: \(13x+4 + 10x-1=23x + 3\).
Then solve the equation \(23x+3 = 141\). Subtract 3 from both sides: \(23x=141 - 3=138\). Divide both sides by 23: \(x = 6\).
\(m\angle PQS=13x + 4=13\times6+4=82^{\circ}\)
\(m\angle SQR=10x-1=10\times6-1 = 59^{\circ}\)
Step4: Use angle - addition postulate for question 4
Since \(m\angle DEF=m\angle DEG + m\angle GEF\), we have the equation \((7x + 4)=(5x + 1)+23\).
Simplify the right - hand side: \((5x + 1)+23=5x+24\).
Then solve the equation \(7x + 4=5x+24\). Subtract \(5x\) from both sides: \(7x-5x+4=5x-5x + 24\), \(2x+4 = 24\). Subtract 4 from both sides: \(2x=20\). Divide both sides by 2: \(x = 10\).
\(m\angle DEG=5x + 1=5\times10+1=51^{\circ}\)
\(m\angle DEF=7x + 4=7\times10+4=74^{\circ}\)
Step5: Use angle - addition postulate for question 5
Since \(m\angle JKL=m\angle JKM + m\angle MKL\), we have the equation \((10x-11)=43+(8x - 20)\).
Simplify the right - hand side: \(43+(8x - 20)=8x+23\).
Then solve the equation \(10x-11=8x + 23\). Subtract \(8x\) from both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both both