Question

1. find (mangle fgh) if (mangle zgh = 128^{circ}) and (mangle fgz=22^{circ}). 12) (mangle jhg = 171^{circ}) and (mangle khg = 60^{circ}). find (mangle jhk). 13) (mangle pqe = 75^{circ}) and (mangle pqr = 165^{circ}). find (mangle eq r). 14) find (mangle cde) if (mangle fde = 110^{circ}) and (mangle cdf = 33^{circ}). 15) (mangle hef = 25^{circ}) and (mangle def = 102^{circ}). find (mangle deh). 16) (mangle aef = 21^{circ}) and (mangle dea = 130^{circ}). find (mangle def). 17) find (mangle klb) if (mangle klm = 179^{circ}) and (mangle blm = 155^{circ}). 18) (mangle thg = 57^{circ}) and (mangle iht = 63^{circ}). find (mangle ihg). 19) (mangle irq = 80^{circ}) and (mangle sri = 26^{circ}). find (mangle srq). 20) (mangle uvw = 162^{circ}) and (mangle uvt = 57^{circ}). find (mangle tvw).

Explanation:

Step1: Recall angle - addition postulate

The measure of the whole angle is the sum of the measures of its non - overlapping parts. For example, if $\angle ABC=\angle ABD+\angle DBC$, then we can use this relationship to solve for unknown angles.

Step2: Solve problem (11)

We know that $\angle FGH = 128^{\circ}$ and $\angle FGE=22^{\circ}$. By the angle - addition postulate, $\angle EGH=\angle FGH-\angle FGE$. So, $\angle EGH = 128^{\circ}-22^{\circ}=106^{\circ}$.

Step3: Solve problem (13)

Given $\angle PQE = 75^{\circ}$ and $\angle PQR = 165^{\circ}$. Then $\angle EQR=\angle PQR - \angle PQE=165^{\circ}-75^{\circ}=90^{\circ}$.

Step4: Solve problem (14)

We have $\angle FDE = 110^{\circ}$ and $\angle CDF = 33^{\circ}$. By the angle - addition postulate, $\angle CDE=\angle FDE-\angle CDF$. So, $\angle CDE = 110^{\circ}-33^{\circ}=77^{\circ}$.

Step5: Solve problem (15)

Given $\angle HEF = 25^{\circ}$ and $\angle DEF = 102^{\circ}$. Then $\angle DEH=\angle DEF-\angle HEF=102^{\circ}-25^{\circ}=77^{\circ}$.

Step6: Solve problem (16)

We know that $\angle AEF = 21^{\circ}$ and $\angle DEA = 130^{\circ}$. By the angle - addition postulate, $\angle DEF=\angle DEA-\angle AEF$. So, $\angle DEF = 130^{\circ}-21^{\circ}=109^{\circ}$.

Step7: Solve problem (17)

Given $\angle KLM = 179^{\circ}$ and $\angle BLM = 155^{\circ}$. Then $\angle KLB=\angle KLM-\angle BLM=179^{\circ}-155^{\circ}=24^{\circ}$.

Step8: Solve problem (18)

We have $\angle THG = 57^{\circ}$ and $\angle IHT = 63^{\circ}$. By the angle - addition postulate, $\angle IHG=\angle THG+\angle IHT$. So, $\angle IHG = 57^{\circ}+63^{\circ}=120^{\circ}$.

Step9: Solve problem (19)

Given $\angle IRQ = 80^{\circ}$ and $\angle SRI = 26^{\circ}$. Then $\angle SRQ=\angle IRQ-\angle SRI=80^{\circ}-26^{\circ}=54^{\circ}$.

Step10: Solve problem (20)

We know that $\angle UVW = 162^{\circ}$ and $\angle UVT = 57^{\circ}$. By the angle - addition postulate, $\angle TVW=\angle UVW-\angle UVT$. So, $\angle TVW = 162^{\circ}-57^{\circ}=105^{\circ}$.

Answer:

(11) $106^{\circ}$
(13) $90^{\circ}$
(14) $77^{\circ}$
(15) $77^{\circ}$
(16) $109^{\circ}$
(17) $24^{\circ}$
(18) $120^{\circ}$
(19) $54^{\circ}$
(20) $105^{\circ}$