Question

1. in the diagram, (mangle agc = 38^{circ}), (mangle cgd = 71^{circ}), and (mangle fgc = 147^{circ}). find each angle measure.

a. (mangle agb) b. (mangle dgf) c. (mangle agf) d. (mangle bgd)

1. smp.1 dig deeper how many times between 12 a.m. and 12 p.m. do the minute hand and hour hand of a clock form a right angle? explain how you found your answer.

Explanation:

Step1: Analyze angle - addition and - subtraction relationships

We know that angles around a point add up to 360°. Also, we use the given angle - measures to find the unknown ones by addition or subtraction.

Step2: Find \(m\angle AGB\)

Since \(\angle AGC\) and \(\angle CGB\) are adjacent angles and we assume \(\angle AGB\) and \(\angle AGC\) are related in a way that if we consider the straight - line property (angles on a straight - line sum to 180°), and no other information about \(\angle B\) and \(\angle C\) position relative to the straight - line is given, we assume \(\angle AGB\) and \(\angle AGC\) are supplementary. But if we consider the non - straight - line case and assume the angles are part of a general angle - relationship around point \(G\), and since no other information about \(\angle B\) is given in relation to other angles, we assume \(\angle AGB = 0^{\circ}\) (if \(B\) lies on \(AG\)). However, if we assume \(B\) is such that \(\angle AGC\) and \(\angle CGB\) are adjacent and \(\angle AGB\) is part of the angle - sum around \(G\), and since no other information is against it, we assume \(\angle AGB\) is supplementary to \(\angle AGC\) in a non - overlapping way. So \(m\angle AGB=180^{\circ}-m\angle AGC = 180 - 38=142^{\circ}\).

Step3: Find \(m\angle DGF\)

We know that \(m\angle FGC=147^{\circ}\) and \(m\angle CGD = 71^{\circ}\). Then \(m\angle DGF=m\angle FGC - m\angle CGD=147 - 71 = 76^{\circ}\).

Step4: Find \(m\angle AGF\)

\(m\angle AGF=m\angle AGC+m\angle FGC=38 + 147=185^{\circ}\) (if we consider the non - overlapping sum of the two angles around \(G\)).

Step5: Find \(m\angle BGD\)

First, we know that the sum of angles around a point is 360°. Let's assume we want to find \(m\angle BGD\) using the known angles. We know \(m\angle AGC = 38^{\circ}\), \(m\angle CGD = 71^{\circ}\), and assume \(m\angle AGB\) as found above. If we consider the angle - sum around \(G\), \(m\angle BGD=360-(m\angle AGB + m\angle AGC+m\angle CGD)\). Substituting \(m\angle AGB = 142^{\circ}\), \(m\angle AGC = 38^{\circ}\), and \(m\angle CGD = 71^{\circ}\), we get \(m\angle BGD=360-(142 + 38+71)=360 - 251 = 109^{\circ}\).

Step6: Solve for the number of right - angle formations of clock hands

The minute hand moves 360° in 60 minutes, so its rate is \(v_m = 6^{\circ}\) per minute. The hour hand moves 360° in 12 hours (720 minutes), so its rate is \(v_h=0.5^{\circ}\) per minute.
Let \(t\) minutes after 12:00. The angle of the minute hand from the 12 - o'clock position is \(\theta_m = 6t\) and the angle of the hour hand from the 12 - o'clock position is \(\theta_h=0.5t\).
The difference between the two angles \(\Delta\theta=\vert\theta_m-\theta_h\vert=\vert6t - 0.5t\vert=\vert5.5t\vert\).
We want \(\Delta\theta = 90\) or \(\Delta\theta = 270\).
For \(\Delta\theta = 90\), we have \(5.5t=90\) or \(5.5t = 270\). Solving \(5.5t=90\), we get \(t=\frac{90}{5.5}=\frac{180}{11}\approx16.36\) minutes and for \(5.5t = 270\), we get \(t=\frac{270}{5.5}=\frac{540}{11}\approx49.09\) minutes.
In 12 hours (720 minutes), the number of times the hands form a right - angle:
The hands form a right - angle approximately every \(\frac{180}{11}\) minutes for one type of right - angle and every \(\frac{540}{11}\) minutes for the other type. The number of times they form a right - angle in 12 hours is 22.

Answer:

a. \(m\angle AGB = 142^{\circ}\)
b. \(m\angle DGF = 76^{\circ}\)
c. \(m\angle AGF = 185^{\circ}\)
d. \(m\angle BGD = 109^{\circ}\)

1. The minute hand and hour hand of a clock form a right - angle 22 times between 12 A.M. and 12 P.M.