QUESTION IMAGE
Question
find the angle measure. then classify the angle. (see example 2.)
- ( mangle bod )
- ( mangle aoe )
- ( mangle coe )
- ( mangle cod )
error analysis describe and correct the error in finding the angle measure in the diagram at the right.
To solve for the angle measures, we use the protractor scale. Let's take \( m\angle COE \) (problem 7) as an example:
Step 1: Identify the rays forming the angle
\( \angle COE \) is formed by rays \( OC \) and \( OE \).
Step 2: Read the protractor scale
- Ray \( OC \) aligns with a mark (e.g., \( 20^\circ \) from \( OA \), but more accurately, observe the protractor: \( OE \) is at \( 60^\circ \) (from \( OB \), but since \( OA \) and \( OB \) are a straight line, we use the inner/outer scale correctly). Wait, let's re-examine:
- \( OA \) is the left horizontal ray (0°), \( OB \) is the right horizontal ray (180°).
- Ray \( OC \): Let's check the protractor. If \( OC \) is at \( 20^\circ \) (from \( OA \), inner scale), and \( OE \) is at \( 60^\circ \) (from \( OA \), inner scale)? Wait, no—wait the protractor has two scales: inner (0° to 180° left to right) and outer (0° to 180° right to left).
For \( \angle COE \):
- \( OC \) is at \( 20^\circ \) (inner scale, from \( OA \) (0°) to \( OC \): 20°).
- \( OE \) is at \( 60^\circ \) (inner scale, from \( OA \) to \( OE \): 60°).
So the measure is \( 60^\circ - 20^\circ = 40^\circ \)? Wait, no—wait the diagram: Let's look at the protractor marks. Wait, the ray \( D \) is at \( 120^\circ \) from \( OB \)? No, let's clarify:
Wait, the protractor is a semicircle, so \( OA \) is 0°, \( OB \) is 180°. The inner scale (bottom) goes from 0° (OA) to 180° (OB) rightward. The outer scale (top) goes from 0° (OB) to 180° (OA) leftward.
Let's take \( \angle COE \):
- \( OC \): Let's see the mark. If \( OC \) is at 20° (inner scale, from OA: 20°).
- \( OE \): Let's see the mark. If \( OE \) is at 60° (inner scale, from OA: 60°).
Then \( m\angle COE = 60^\circ - 20^\circ = 40^\circ \). Wait, but maybe I misread. Alternatively, if \( OC \) is at 160° (outer scale, from OB: 160°), and \( OE \) is at 120° (outer scale, from OB: 120°), then \( 160^\circ - 120^\circ = 40^\circ \). Either way, \( \angle COE \) is \( 40^\circ \), which is an acute angle (less than 90°).
For \( m\angle BOD \) (problem 5):
- \( OB \) is 180° (or 0° on outer scale). \( OD \) is at 60° from \( OB \) (outer scale: 180° - 120° = 60°? Wait, the ray \( D \) is at 120° from \( OA \) (inner scale: 120°). So \( m\angle BOD = 180^\circ - 120^\circ = 60^\circ \)? No, wait: \( OA \) is 0°, \( OB \) is 180°. If \( OD \) is at 120° from \( OA \) (inner scale), then from \( OB \) (180°), it's \( 180^\circ - 120^\circ = 60^\circ \). So \( m\angle BOD = 60^\circ \), acute.
For \( m\angle AOE \) (problem 6):
- \( OA \) is 0°, \( OE \) is at 60° (inner scale) from \( OA \)? Wait, no—if \( OE \) is at 60° from \( OB \) (outer scale: 60°), then from \( OA \) (0°), it's \( 180^\circ - 60^\circ = 120^\circ \)? Wait, confusion between inner and outer scales. Let's use the correct method: the angle between \( OA \) (left horizontal) and \( OE \). If \( OE \) is at 60° above the right horizontal (\( OB \)), then from \( OA \) (left horizontal), it's \( 180^\circ - 60^\circ = 120^\circ \). So \( m\angle AOE = 120^\circ \), obtuse (between 90° and 180°).
For \( m\angle COD \) (problem 8):
- \( OC \) is at 20° from \( OA \) (inner scale), \( OD \) is at 120° from \( OA \) (inner scale). So \( m\angle COD = 120^\circ - 20^\circ = 100^\circ \)? Wait, no—if \( OC \) is at 20° (inner) and \( OD \) is at 120° (inner), then \( 120 - 20 = 100^\circ \), obtuse.
Error Analysis (if applicable):
Common error: Using the wrong protractor scale (inner vs. outer). For example, if finding \( \angle AOE \), someone…
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To solve for the angle measures, we use the protractor scale. Let's take \( m\angle COE \) (problem 7) as an example:
Step 1: Identify the rays forming the angle
\( \angle COE \) is formed by rays \( OC \) and \( OE \).
Step 2: Read the protractor scale
- Ray \( OC \) aligns with a mark (e.g., \( 20^\circ \) from \( OA \), but more accurately, observe the protractor: \( OE \) is at \( 60^\circ \) (from \( OB \), but since \( OA \) and \( OB \) are a straight line, we use the inner/outer scale correctly). Wait, let's re-examine:
- \( OA \) is the left horizontal ray (0°), \( OB \) is the right horizontal ray (180°).
- Ray \( OC \): Let's check the protractor. If \( OC \) is at \( 20^\circ \) (from \( OA \), inner scale), and \( OE \) is at \( 60^\circ \) (from \( OA \), inner scale)? Wait, no—wait the protractor has two scales: inner (0° to 180° left to right) and outer (0° to 180° right to left).
For \( \angle COE \):
- \( OC \) is at \( 20^\circ \) (inner scale, from \( OA \) (0°) to \( OC \): 20°).
- \( OE \) is at \( 60^\circ \) (inner scale, from \( OA \) to \( OE \): 60°).
So the measure is \( 60^\circ - 20^\circ = 40^\circ \)? Wait, no—wait the diagram: Let's look at the protractor marks. Wait, the ray \( D \) is at \( 120^\circ \) from \( OB \)? No, let's clarify:
Wait, the protractor is a semicircle, so \( OA \) is 0°, \( OB \) is 180°. The inner scale (bottom) goes from 0° (OA) to 180° (OB) rightward. The outer scale (top) goes from 0° (OB) to 180° (OA) leftward.
Let's take \( \angle COE \):
- \( OC \): Let's see the mark. If \( OC \) is at 20° (inner scale, from OA: 20°).
- \( OE \): Let's see the mark. If \( OE \) is at 60° (inner scale, from OA: 60°).
Then \( m\angle COE = 60^\circ - 20^\circ = 40^\circ \). Wait, but maybe I misread. Alternatively, if \( OC \) is at 160° (outer scale, from OB: 160°), and \( OE \) is at 120° (outer scale, from OB: 120°), then \( 160^\circ - 120^\circ = 40^\circ \). Either way, \( \angle COE \) is \( 40^\circ \), which is an acute angle (less than 90°).
For \( m\angle BOD \) (problem 5):
- \( OB \) is 180° (or 0° on outer scale). \( OD \) is at 60° from \( OB \) (outer scale: 180° - 120° = 60°? Wait, the ray \( D \) is at 120° from \( OA \) (inner scale: 120°). So \( m\angle BOD = 180^\circ - 120^\circ = 60^\circ \)? No, wait: \( OA \) is 0°, \( OB \) is 180°. If \( OD \) is at 120° from \( OA \) (inner scale), then from \( OB \) (180°), it's \( 180^\circ - 120^\circ = 60^\circ \). So \( m\angle BOD = 60^\circ \), acute.
For \( m\angle AOE \) (problem 6):
- \( OA \) is 0°, \( OE \) is at 60° (inner scale) from \( OA \)? Wait, no—if \( OE \) is at 60° from \( OB \) (outer scale: 60°), then from \( OA \) (0°), it's \( 180^\circ - 60^\circ = 120^\circ \)? Wait, confusion between inner and outer scales. Let's use the correct method: the angle between \( OA \) (left horizontal) and \( OE \). If \( OE \) is at 60° above the right horizontal (\( OB \)), then from \( OA \) (left horizontal), it's \( 180^\circ - 60^\circ = 120^\circ \). So \( m\angle AOE = 120^\circ \), obtuse (between 90° and 180°).
For \( m\angle COD \) (problem 8):
- \( OC \) is at 20° from \( OA \) (inner scale), \( OD \) is at 120° from \( OA \) (inner scale). So \( m\angle COD = 120^\circ - 20^\circ = 100^\circ \)? Wait, no—if \( OC \) is at 20° (inner) and \( OD \) is at 120° (inner), then \( 120 - 20 = 100^\circ \), obtuse.
Error Analysis (if applicable):
Common error: Using the wrong protractor scale (inner vs. outer). For example, if finding \( \angle AOE \), someone might use the outer scale (from \( OB \)) as 60° and think it's 60°, but actually, from \( OA \), it's \( 180 - 60 = 120^\circ \). To correct, identify the correct scale (inner for angles opening from \( OA \) to the right, outer for angles opening from \( OB \) to the left) and subtract/add appropriately.
Final Answers (example for \( m\angle COE \)):
\( m\angle COE = 40^\circ \), and it is an acute angle (since \( 0^\circ < 40^\circ < 90^\circ \)).
(Adjust based on the exact protractor marks, but the process is: identify the two rays, read their positions on the protractor, subtract to find the angle, then classify as acute (0–90°), right (90°), obtuse (90–180°), straight (180°), or reflex (180–360°).)