Question

1. m∠abc =
2. if m∠abd = 127°, then m∠abc =
3. if m∠xyz = 78°, then m∠wyz =
4. m∠ops =
5. if m∠def = 150°, then m∠cef =
6. if m∠lmo = 122°, then m∠lmp =

answer: 4 9 2 13 14 7 0 10 9 6 15 9 9 5 3 9 6 9 7 5 1 11 8 12

Explanation:

Step1: Find $\angle ABC$ for first problem

Sum of angles at point $B$ for first - figure: $\angle ABC=46^{\circ}+58^{\circ}=104^{\circ}$

Step2: Find $\angle ABC$ for second problem

Since $\angle ABD = 127^{\circ}$ and $\angle DBC = 49^{\circ}$, then $\angle ABC=\angle ABD-\angle DBC = 127^{\circ}-49^{\circ}=78^{\circ}$

Step3: Find $\angle WYZ$ for third problem

$\angle WYZ=\angle XYZ - \angle XYW=78^{\circ}-21^{\circ}=57^{\circ}$

Step4: Find $\angle QPS$ for fourth problem

Since $\angle QPS$ is a right - angle, $(6x - 4)^{\circ}+(7x + 16)^{\circ}=90^{\circ}$. Combine like terms: $13x+12 = 90$, $13x=78$, $x = 6$. Then $\angle QPS=(6x - 4)^{\circ}+(7x + 16)^{\circ}=(6\times6 - 4)+(7\times6 + 16)=32 + 58=90^{\circ}$

Step5: Find $\angle CEP$ for fifth problem

Since $\angle DEF = 150^{\circ}$, and $\angle DEF=(9x + 4)^{\circ}+(5x + 15)^{\circ}$, then $14x+19 = 150$, $14x = 131$, $x=\frac{131}{14}\approx9.36$. $\angle CEP=(5x + 15)^{\circ}=5\times9.36+15=46.8 + 15=61.8^{\circ}\approx62^{\circ}$

Step6: Find $\angle LMP$ for sixth problem

Since $\angle LMO = 122^{\circ}$, and $\angle LMO=(4x - 19)^{\circ}+(x + 60)^{\circ}$, then $5x + 41=122$, $5x=81$, $x=\frac{81}{5}=16.2$. $\angle LMP=(4x - 19)^{\circ}=4\times16.2-19=64.8-19 = 45.8^{\circ}\approx46^{\circ}$

Answer:

1. $104^{\circ}$
2. $78^{\circ}$
3. $57^{\circ}$
4. $90^{\circ}$
5. $62^{\circ}$
6. $46^{\circ}$