Question

what do you get when you,,,1. cross two ducks with a match?answer: \frac{f}{37^{\circ}} \frac{i}{57^{\circ}} \frac{}{99^{\circ}} \frac{e}{67^{\circ}} \frac{}{104^{\circ}} \frac{}{76^{\circ}} \frac{u}{59^{\circ}} \frac{a}{113^{\circ}} \frac{c}{42^{\circ}} \frac{k}{53^{\circ}} \frac{e}{67^{\circ}} \frac{}{99^{\circ}} \frac{s}{18^{\circ}}2. cross a stick of dynamite with a lemon pie?answer: \frac{a}{113^{\circ}} \frac{}{68^{\circ}} \frac{}{63^{\circ}} \frac{o}{34^{\circ}} \frac{o}{34^{\circ}} \frac{}{54^{\circ}} \frac{}{38^{\circ}} \frac{}{54^{\circ}} \frac{f}{67^{\circ}} \frac{}{99^{\circ}} \frac{i}{57^{\circ}} \frac{n}{90^{\circ}} \frac{g}{36^{\circ}} \frac{u}{59^{\circ}} \frac{e}{67^{\circ}}find the angle measures indicated. look for each answer in the code. each time theanswer appears, write the letter of the exercise above it.\textcircled{i} m \angle b = 57^{\circ} \textcircled{g} m \angle j = 36^{\circ} \textcircled{s} m \angle wox = 18^{\circ}\textcircled{a} m \angle pqr = 113^{\circ} \textcircled{n} m \angle dab = 90^{\circ} \textcircled{c} m \angle xzy = 42^{\circ}\textcircled{e} m \angle pqt = 67^{\circ} \textcircled{o} m \angle dac = 34^{\circ} \textcircled{u} m \angle y = 59^{\circ}\textcircled{f} m \angle mnl = 37^{\circ} \textcircled{q} m \angle efd = \textcircled{m} m \angle aob =\textcircled{k} m \angle m = 53^{\circ} \textcircled{b} m \angle e = \textcircled{r} m \angle boc =middle school math with pizzazz! book d\textcopyright creative publications d - 35 topic 3 - k: review: related angles

Explanation:

Step1: Find \(m\angle EFD\)

Since \( \angle EFD\) and \(104^{\circ}\) are vertical angles. Vertical angles are equal. So \(m\angle EFD = 104^{\circ}\)

Step2: Find \(m\angle E\)

In the triangle with angles \(41^{\circ}\), \( \angle E\) and \( \angle EFD\). Using the triangle - angle sum theorem (\(m\angle A+m\angle B + m\angle C=180^{\circ}\)). Let \(m\angle A = 41^{\circ}\), \(m\angle C=m\angle EFD = 104^{\circ}\). Then \(m\angle E=180-(41 + 104)=35^{\circ}\)

Step3: Find \(m\angle AOB\)

\(\angle AOB\) and \(54^{\circ}\) are vertical angles. Vertical angles are equal. So \(m\angle AOB = 54^{\circ}\)

Step4: Find \(m\angle BOC\)

\(\angle AOB+\angle BOC+\angle COD+\angle DOA = 360^{\circ}\). Since \(\angle AOD = 27^{\circ}\), \(\angle AOB = 54^{\circ}\), \(\angle COD\) and \(\angle AOB\) are vertical angles (\(\angle COD = 54^{\circ}\)). Then \(m\angle BOC=360-(27 + 54+54)=225^{\circ}\) (This part may be wrong as there is a mis - understanding of the figure. If we consider the straight - line property: \(\angle AOB+\angle BOC+\angle COD+\angle DOA = 360^{\circ}\). If we assume the figure is formed by two intersecting lines. Another way: \(\angle AOB\) and \(\angle BOC\) are adjacent angles. If we consider the sum of angles around a point. But if we consider the fact that \(\angle AOB\) and \(27^{\circ}\) and \(\angle BOC\) and \(54^{\circ}\) are related. Using the property that the sum of angles around a point \(O\) is \(360^{\circ}\). Also, if we consider the fact that \(\angle AOB\) and \(\angle COD\) are vertical angles (\(\angle AOB=\angle COD = 54^{\circ}\)), \(\angle AOD\) and \(\angle BOC\) are vertical angles. So \(m\angle BOC = 27^{\circ}\))

Answer:

\(m\angle EFD = 104^{\circ}\), \(m\angle E = 35^{\circ}\), \(m\angle AOB = 54^{\circ}\), \(m\angle BOC = 27^{\circ}\)