7.3 Arcs and Central Angles Cung và góc ở tâm

The major arc between $A$ and $B$ measures $290^\circ,$ and a major arc measures $360^\circ$ minus its minor arc. $\angle 1$ is the central angle of the minor arc $\overset{\frown}{AB}$:

Answer: $m\angle 1 = 70^\circ.$

$\angle DOB$ is the central angle of $\overset{\frown}{DB},$ so $m\angle DOB = 40^\circ.$ $\overline{OC} \perp \overline{AB},$ so $m\angle COB = 90^\circ,$ and $\angle 1 = \angle COD$ is the rest of it:

Answer: $m\angle 1 = 50^\circ.$

The 12 hour marks cut the clock face into $12$ equal arcs, each $\dfrac{360^\circ}{12} = 30^\circ.$ At five o'clock the minute hand points to 12 and the hour hand to 5, five arcs away:

Answer: $150^\circ.$

At 2:00 the hands are two arcs of $30^\circ$ apart: $2 \cdot 30^\circ = 60^\circ.$

In five minutes the minute hand turns $5 \cdot 6^\circ = 30^\circ,$ but the hour hand also moves: it turns $30^\circ$ in $60$ minutes, so $0.5^\circ$ a minute. Measured from 12:

The angle is not $60^\circ - 30^\circ = 30^\circ,$ because the hour hand has moved $2.5^\circ$ away from 2.

Answer: the angle at 2:05 is $32.5^\circ,$ not $30^\circ.$

$OR = OT$ (radii), so $\triangle ORT$ is isosceles and $m\angle ORT = m\angle 1.$ $\angle ROS$ is an exterior angle of $\triangle ORT,$ so it equals the sum of the two remote interior angles; it is also the central angle of $\overset{\frown}{RS}$:

a. $m\overset{\frown}{RS} = 2 \cdot 28^\circ = 56^\circ.$

b. $m\angle 1 = \dfrac{m\overset{\frown}{RS}}{2} = \dfrac{76^\circ}{2} = 38^\circ.$

Answer: a. $56^\circ$; b. $38^\circ.$

Here $\angle C$ is $\angle ACB.$ $PA = PB = PC$ (radii), so $\triangle PCB$ and $\triangle PCA$ are isosceles.

a. $m\angle CPB = m\overset{\frown}{BC} = 140^\circ$ (central angle), and $\angle APC$ and $\angle CPB$ form a linear pair, so $m\angle APC = 40^\circ.$ The base angles of each isosceles triangle are equal:

b. The same steps with $n$:

Answer: a. $90^\circ$; b. $90^\circ$ for every $n.$

Given: in $\odot O,$ $\overset{\frown}{AB} = \overset{\frown}{CD}$ (minor arcs). Prove: $\angle AOB = \angle COD.$

1. $\overset{\frown}{AB} = \overset{\frown}{CD}$ (given), so $m\overset{\frown}{AB} = m\overset{\frown}{CD}$ by the definition of equal arcs.

2. $m\overset{\frown}{AB} = m\angle AOB$ and $m\overset{\frown}{CD} = m\angle COD,$ by the definition of the measure of a minor arc.

3. So $m\angle AOB = m\angle COD$ by substitution, and $\angle AOB = \angle COD.$

Claim: $\overset{\frown}{YZ} = \overset{\frown}{ZW}.$ Draw radius $\overline{OY}.$

1. $OX = OY$ (radii), so $\angle OXY = \angle OYX$ (base angles of an isosceles triangle).

2. $\overline{XY} \parallel \overline{OZ}$ with transversal $\overline{XW}$: $\angle ZOW = \angle OXY$ (corresponding angles).

3. $\overline{XY} \parallel \overline{OZ}$ with transversal $\overline{OY}$: $\angle YOZ = \angle OYX$ (alternate interior angles).

4. From 1, 2 and 3: $\angle YOZ = \angle ZOW$ (substitution).

5. A minor arc measures its central angle, so $m\overset{\frown}{YZ} = m\overset{\frown}{ZW}$: $\overset{\frown}{YZ} = \overset{\frown}{ZW}$ (equal arcs of one circle).