Question

topic: date: main ideas/questions class: notes/examples angle addition postulate if d is in the interior of ∠abc, then ____________________ examples use the diagram below to answer questions 1 and 2. 1. if m∠abd = 48° and m∠dbc = 78°, find m∠abc. 2. if m∠dbc = 74° and m∠abc = 119°, find m∠abd. 3. if m∠pqr = 141°, find each measure. x = ____ m∠pqs = ____ m∠sqr = ____ 4. if m∠def=(7x + 4)°, m∠deg=(5x + 1)°, and m∠gef = 23°, find each measure. x = ____ m∠deg = ____ m∠def = ____ 5. if m∠jkm = 43°, m∠mkl=(8x - 20)°, and m∠jkl=(10x - 11)°, find each measure. x = ____ m∠mkl = ____ m∠jkl = ______ © gina wilson (all things algebra®, llc). 2014 - 2019

Explanation:

Step1: Recall angle - addition postulate

If a point $D$ is in the interior of $\angle ABC$, then $m\angle ABC=m\angle ABD + m\angle DBC$.

Step2: Solve question 1

Given $m\angle ABD = 48^{\circ}$ and $m\angle DBC=78^{\circ}$. By the angle - addition postulate, $m\angle ABC=m\angle ABD + m\angle DBC=48^{\circ}+78^{\circ}=126^{\circ}$.

Step3: Solve question 2

Given $m\angle DBC = 74^{\circ}$ and $m\angle ABC = 119^{\circ}$. Using the angle - addition postulate $m\angle ABC=m\angle ABD + m\angle DBC$, we can find $m\angle ABD=m\angle ABC - m\angle DBC=119^{\circ}-74^{\circ}=45^{\circ}$.

Step4: Solve question 3

Since $m\angle PQR=m\angle PQS + m\angle SQR$ and $m\angle PQR = 141^{\circ}$, $m\angle PQS=(13x + 4)^{\circ}$, $m\angle SQR=(10x-1)^{\circ}$. Then $(13x + 4)+(10x-1)=141$. Combine like - terms: $23x+3 = 141$. Subtract 3 from both sides: $23x=138$. Divide both sides by 23: $x = 6$. So $m\angle PQS=13x + 4=13\times6 + 4=82^{\circ}$, $m\angle SQR=10x-1=10\times6-1 = 59^{\circ}$.

Step5: Solve question 4

Since $m\angle DEF=m\angle DEG + m\angle GEF$, and $m\angle DEF=(7x + 4)^{\circ}$, $m\angle DEG=(5x + 1)^{\circ}$, $m\angle GEF = 23^{\circ}$. Then $(5x + 1)+23=7x + 4$. Combine like - terms: $5x+24=7x + 4$. Subtract $5x$ from both sides: $24=2x + 4$. Subtract 4 from both sides: $20=2x$. Divide both sides by 2: $x = 10$. So $m\angle DEG=5x + 1=5\times10+1 = 51^{\circ}$, $m\angle DEF=7x + 4=7\times10+4 = 74^{\circ}$.

Step6: Solve question 5

Since $m\angle JKL=m\angle JKM + m\angle MKL$, and $m\angle JKM = 43^{\circ}$, $m\angle MKL=(8x - 20)^{\circ}$, $m\angle JKL=(10x-11)^{\circ}$. Then $43+(8x - 20)=10x-11$. Combine like - terms: $23 + 8x=10x-11$. Subtract $8x$ from both sides: $23=2x-11$. Add 11 to both sides: $34=2x$. Divide both sides by 2: $x = 17$. So $m\angle MKL=8x-20=8\times17-20 = 116^{\circ}$, $m\angle JKL=10x-11=10\times17-11 = 159^{\circ}$.

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. $x = 17$, $m\angle MKL = 116^{\circ}$, $m\angle JKL = 159^{\circ}$