Question

use the diagram below to answer questions 1 and 2.

1. if (mangle abd = 48^{circ}) and (mangle dbc=78^{circ}), find (mangle abc).
2. if (mangle dbc = 74^{circ}) and (mangle abc = 119^{circ}), find (mangle abd).
3. if (mangle pqr = 141^{circ}), find each measure.

• ((13x + 4)^{circ}) and ((10x - 1)^{circ})
• (x=)
• (mangle pqs=)
• (mangle sqr=)

1. if (mangle def=(7x + 4)^{circ}), (mangle deg=(5x + 1)^{circ}), and (mangle gef = 23^{circ}), find each measure.

• (x=)
• (mangle deg=)
• (mangle def=)

1. if (mangle jkm = 43^{circ}), (mangle mkl=(8x - 20)^{circ}), and (mangle jkl=(10x - 11)^{circ}), find each measure.

• (x=)
• (mangle mkl=)
• (mangle jkl=)

Explanation:

Step1: Use angle - addition postulate for question 1

Since $\angle ABC=\angle ABD+\angle DBC$, substituting $m\angle ABD = 48^{\circ}$ and $m\angle DBC = 78^{\circ}$, we get $m\angle ABC=48^{\circ}+78^{\circ}=126^{\circ}$.

Step2: Use angle - addition postulate for question 2

Since $\angle ABC=\angle ABD+\angle DBC$, then $\angle ABD=\angle ABC - \angle DBC$. Substituting $m\angle DBC = 74^{\circ}$ and $m\angle ABC = 119^{\circ}$, we have $m\angle ABD=119^{\circ}-74^{\circ}=45^{\circ}$.

Step3: Use angle - addition postulate for question 3

Since $\angle PQR=\angle PQS+\angle SQR$, and $\angle PQS=(13x + 4)^{\circ}$, $\angle SQR=(10x-1)^{\circ}$, $\angle PQR = 141^{\circ}$. Then $(13x + 4)+(10x-1)=141$.
Combining like - terms: $13x+10x+4 - 1=141$, $23x+3 = 141$.
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 $\angle DEF=\angle DEG+\angle GEF$, then $(7x + 4)=(5x + 1)+23$.
Expand the right - hand side: $7x+4=5x+24$.
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 $\angle JKL=\angle JKM+\angle MKL$, then $(10x-11)=43+(8x - 20)$.
Expand the right - hand side: $10x-11=8x+23$.
Subtract $8x$ from both sides: $10x-8x-11=8x-8x + 23$, $2x-11=23$.
Add 11 to both sides: $2x=34$.
Divide both sides by 2: $x = 17$.
$m\angle MKL=8x-20=8\times17-20=116^{\circ}$.
$m\angle JKL=10x-11=10\times17-11=159^{\circ}$.

Answer:

1. $m\angle ABC = 126^{\circ}$
2. $m\angle ABD = 45^{\circ}$
3. $x = 6$, $m\angle PQS = 82^{\circ}$, $m\angle SQR = 59^{\circ}$
4. $x = 10$, $m\angle DEG = 51^{\circ}$, $m\angle DEF = 74^{\circ}$
5. $x = 17$, $m\angle MKL = 116^{\circ}$, $m\angle JKL = 159^{\circ}$