Question

use the protractor to measure the angle. ∠ abc = \square °

Explanation:

Step1: Align protractor with angle

Place protractor center at \( B \), base line along \( BA \).

Step2: Read angle for \( BC \)

The ray \( BC \) aligns with \( 30^\circ \) on the protractor (since the other side is \( 180^\circ - 30^\circ = 150^\circ \), but the angle here is the smaller one? Wait, no—wait, the angle \( \angle ABC \): \( BA \) is along the 0° (or 180°) line, and \( BC \) is at 30° from \( BA \) but on the other side? Wait, no, looking at the protractor: the bottom line is \( BA \) (along 0° and 180°), and \( BC \) is at 30° above? Wait, no, the angle between \( BA \) and \( BC \): let's check the protractor. The protractor has two scales: inner and outer. The outer scale (from 0° to 180° clockwise) and inner (counter-clockwise). \( BA \) is along the 0° (outer) and 180° (inner). \( BC \) is at 30° on the outer scale? Wait, no, the ray \( BC \) is at 30° from \( BA \) but in the other direction? Wait, no, the angle \( \angle ABC \): the vertex is \( B \), sides \( BA \) and \( BC \). So \( BA \) is along the horizontal line (0°), and \( BC \) is at 30° above? Wait, no, looking at the protractor, the mark for \( BC \) is at 30°? Wait, no, the protractor shows that the angle between \( BA \) (horizontal) and \( BC \) is 150°? Wait, no, wait: if \( BA \) is along 0° (right side, 0°), and \( BC \) is at 150° from \( BA \) (since 180° - 30° = 150°). Wait, no, let's re-examine. The protractor's outer scale: 0° at the right (along \( BA \)), and moving counter-clockwise, the angle for \( BC \) is 150°? Wait, no, the ray \( BC \) is at 30° on the inner scale? Wait, maybe I made a mistake. Wait, the angle \( \angle ABC \): \( BA \) is the horizontal line (from \( B \) to the right, 0°), and \( BC \) is going up to the left, making an angle. Let's check the protractor: the inner scale (counter-clockwise) has 0° at the left (180° outer), but \( BA \) is at the right (0° outer). So the outer scale: 0° at right, 180° at left. The ray \( BC \) is at 30° from the left? No, wait, the protractor's markings: the top scale (outer) goes from 0° (right) to 180° (left), and the bottom scale (inner) goes from 0° (left) to 180° (right). So \( BA \) is along the right (0° outer), and \( BC \) is at 30° on the outer scale? No, that can't be. Wait, the angle between \( BA \) (right, 0°) and \( BC \) (left-up) is 150°, because 180° - 30° = 150°. Wait, no, let's count the degrees. If \( BA \) is at 0°, and \( BC \) is at 30° from the left (180°), then the angle between them is 180° - 30° = 150°? Wait, no, that's not right. Wait, the correct way: place the protractor so that the center is at \( B \), the 0° line along \( BA \). Then read the angle where \( BC \) crosses the protractor. The protractor shows that \( BC \) is at 30° from the 180° line (left), so from \( BA \) (0° right), the angle is 180° - 30° = 150°? Wait, no, I think I messed up. Wait, the angle \( \angle ABC \): \( BA \) is going to the left? No, \( BA \) is going to the left? Wait, the diagram: \( B \) is the vertex, \( BA \) is going down-left, and \( BC \) is going up-left? Wait, no, the horizontal line is \( BA \) (from \( B \) to the right, 0°), and \( BC \) is going up to the left, making an angle. Wait, the protractor's outer scale (clockwise from 0° at right) shows that \( BC \) is at 150°? Wait, no, the protractor's markings: the outer scale (top) has 0° at right, 180° at left. The ray \( BC \) is at 30° from the left (180°), so 180° - 30° = 150°? Wait, no, that's the angle. Wait, maybe the correct angle is 150°? Wait, no, let's check again. The angle between \( BA \) (horizontal right) and \( BC \) (going up to the left) is 150°, because 180° - 30° = 150°. Wait, but the protractor's inner scale (counter-clockwise) would show 30°? No, inner scale is 0° at left (180° outer), so 0° at left, 180° at right. So \( BA \) is at 180° inner (right), and \( BC \) is at 30° inner (counter-clockwise from left). So the angle between \( BA \) (180° inner) and \( BC \) (30° inner) is 180° - 30° = 150°? Yes, that makes sense. So \( \angle ABC = 150^\circ \). Wait, but maybe I made a mistake. Wait, the other way: if \( BA \) is along 0° outer (right), and \( BC \) is at 30° outer (clockwise), but that would be 30°, but that's not possible. Wait, no, the ray \( BC \) is above the horizontal line? No, in the diagram, \( BA \) is going down to the left, and \( BC \) is going up to the left, so the angle between them is 150°. Wait, maybe the correct answer is 150. Wait, let's see: the protractor's outer scale (0° at right, 180° at left) – \( BC \) is at 30° from the left (180°), so 180° - 30° = 150°. So the angle \( \angle ABC \) is 150 degrees.

Step3: Confirm the angle

Check the protractor alignment: center at \( B \), base line along \( BA \), read the angle for \( BC \). The correct reading is 150° (since the angle between \( BA \) and \( BC \) is obtuse, more than 90°).

Answer:

\( 150 \)