Question

1. look for relationships in the diagram below what are the measures of ∠dbc, ∠dbe, ∠ebf, and ∠dbf?

Explanation:

Step1: Measure ∠DBC

∠DBC is a straight - line related angle? No, looking at the protractor, the ray BD and ray BC. The straight line is AC, so ∠ABC is 180°. The angle of BD with BA: from the protractor, BD is at 40° from BA (since the protractor has markings, and BA is the left - most ray). Wait, actually, to find ∠DBC, we can see that the angle between BD and BC. Since ∠ABC = 180°, and the angle between BD and BA is 40° (from the protractor, the lower scale: BA is 180°/0°, BD is at 40° from BA towards BC? Wait, no. The protractor is a semicircle, with center at B. The ray BA is at 180° (left) and BC is at 0° (right). So the angle between BA and BC is 180°. Now, ray BD: looking at the protractor, the lower scale (the one with 0 - 180° from left to right). The mark for BD: the line BD passes through the 40° mark on the lower scale (since from BA (180°) moving towards BC (0°), the angle of BD with BA is 40°? Wait, no. Wait, the protractor has two scales: one going clockwise (from 0° at BC to 180° at BA) and one counter - clockwise (from 0° at BA to 180° at BC). Wait, in the diagram, the protractor is placed with center at B, and the straight edge along AC (BA to BC). So for ∠DBC: the angle between BD and BC. Since BC is at 0° (right), and BD is at 180°−40° = 140°? Wait, no. Wait, let's look at the upper scale. Wait, the upper scale: the numbers are 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180 on the left and 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180 on the right? No, the protractor has a semicircle, so from left (BA) to right (BC), the lower scale is 0° (at BC) to 180° (at BA), and the upper scale is 180° (at BA) to 0° (at BC). Wait, maybe a better way: ∠DBC is the angle between BD and BC. Since BC is along the positive x - axis (right), and BD is at an angle of 180°−40° = 140° from BC? Wait, no. Wait, looking at the diagram, the ray BD: the lower scale (the one with 0 at BC) – no, the lower scale has 0 at BA? Wait, I think I made a mistake. Let's re - examine. The protractor is a semicircular protractor, with center at B. The line AC is a straight line, with A on the left and C on the right. So BA is the left - most ray, BC is the right - most ray. The angle between BA and BC is 180°. Now, the ray BD: when we look at the protractor, the mark for BD on the lower scale (the scale that goes from BA (left) to BC (right), 0° at BC and 180° at BA) – no, the lower scale is from 0° (BC) to 180° (BA) clockwise. The upper scale is from 180° (BA) to 0° (BC) counter - clockwise. Wait, the ray BD: the line BD passes through the 40° mark on the lower scale? No, the lower scale: at BC (right) it's 0°, then moving towards BA (left), it's 10°, 20°, 30°, 40°, 50°,... 180°. Wait, the ray BD: the angle between BD and BA is 40° (since the mark on the lower scale at BD is 40° from BA). So ∠ABD = 40°, then ∠DBC=180°−∠ABD = 180°−40° = 140°.

Step2: Measure ∠DBE

∠DBE is the angle between BD and BE. Let's find the angle between BD and BE. From the protractor, the ray BE: looking at the upper scale, the mark for BE is at 110°? Wait, no. Wait, the upper scale: the numbers on the upper part. The ray BE: the upper scale (the one that goes from 0° at BC to 180° at BA counter - clockwise). Wait, the angle between BD and BE: ∠DBE. Since ∠ABD = 40°, and ∠ABE: from the protractor, the angle between BA and BE. Let's see, the ray BE: the upper scale mark. The upper scale at BE: the number is 110°? Wait, no. Wait, the lower scale: BA is 180°, BC is 0°. The ray BE: looking at the lower scale, the angle between BA and BE is 180°−110° = 70°? Wait, maybe I should use the fact that ∠DBE=∠DBC−∠EBC? No, wait, let's look at the protractor markings. The ray BD is at 40° from BA (∠ABD = 40°), and the ray BE is at 180°−110° = 70° from BA (∠ABE = 70°). So ∠DBE=∠ABE−∠ABD = 70°−40°? No, that can't be. Wait, no, if BA is the left - most, then ∠ABD is 40° (BD is 40° above BA? No, BD is a ray going up - left? Wait, the diagram shows BD as a ray going to the left - upper, BE to the upper - middle, and BF to the upper - right. Wait, maybe the protractor is used to measure the angles directly. Let's assume that:

- ∠DBC: the angle between BD and BC. Since BC is a straight line to the right, and BD is at an angle of 140° (because 180°−40° = 140°).
- ∠DBE: the angle between BD and BE. From the protractor, the difference between the angles of BD and BE from BC. Wait, BD is at 140° from BC (since ∠DBC = 140°), and BE: let's see, the ray BE is at 180°−110° = 70° from BA, so from BC, it's 180°−70° = 110°? No, that's not right. Wait, maybe the protractor has the following: the angle between BD and BC is 140° (∠DBC = 140°), the angle between BD and BE: BD is at 40° from BA, BE is at 110° from BC? No, I think I made a mistake. Let's start over.

The protractor is a semicircle with center at B. The straight line AC (BA to BC) is the diameter. So:

- ∠DBC: To find ∠DBC, we look at the angle between BD and BC. Since BC is at 0° (right), and BD is at 180°−40° = 140° (because the mark for BD on the protractor's lower scale (from BC to BA, 0° to 180°) is 140°? Wait, the lower scale: BC is 0°, BA is 180°. The ray BD passes through the 40° mark on the upper scale? No, the upper scale is from BA (180°) to BC (0°), so the number at BD on the upper scale is 40°? No, the upper scale numbers are 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180 on the left and 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180 on the right? No, the protractor has a semicircle, so the angle between BD and BC: if we use the outer scale (the one with 0° at BC and 180° at BA), the ray BD is at 140° (because 180°−40° = 140°).

- ∠DBE: The angle between BD and BE. The ray BE is at 180°−110° = 70° from BA? No, wait, the ray BE: looking at the protractor, the angle between BA and BE is 70° (because the upper scale at BE is 110° from BC? No, I think the correct way is:

From the protractor, we can see that:

- ∠DBC: The angle between BD and BC. Since BC is a straight line (180° - 40° = 140°), so ∠DBC = 140°.
- ∠DBE: The angle between BD and BE. The angle between BD and BA is 40°, and the angle between BE and BA is 70° (because 180° - 110° = 70°), so ∠DBE=70° - 40°? No, that would be 30°, but that's not right. Wait, no, if BD is 40° from BA (towards BC), and BE is 110° from BC (towards BA), then ∠DBE = ∠DBC - ∠EBC. ∠DBC = 140°, ∠EBC: the angle between BE and BC. From the protractor, the angle between BE and BC is 70° (because 180° - 110° = 70°), so ∠DBE=140° - 70° = 70°? Wait, that makes sense.

- ∠EBF: The angle between BE and BF. From the protractor, the angle between BE and BC is 70°, and the angle between BF and BC is 40° (since BF is at 40° from BC towards BA). Wait, no, the ray BF: looking at the protractor, the mark for BF is at 40° from BC (lower scale). So ∠EBF=∠EBC - ∠FBC = 70° - 40° = 30°?

- ∠DBF: The angle between BD and BF. ∠DBF=∠DBC - ∠FBC = 140° - 40° = 100°? Wait, no, ∠DBF can also be ∠DBE+∠EBF = 70°+30° = 100°.

Wait, let's check with the protractor markings:

- ∠DBC: The angle between BD and BC. Since BC is 0° (right), and BD is at 140° (from the protractor's lower scale, 180° - 40° = 140°), so ∠DBC = 140°.
- ∠DBE: The angle between BD and BE. BD is at 140° from BC, BE is at 110° from BC (upper scale), so 140° - 110° = 30°? No, that's not. Wait, I think I messed up the scales. Let's use the correct method for a protractor:

1. ∠DBC:
- The vertex is B, sides are BD and BC.
- Place the protractor's center at B, align BC with the 0° mark (right - hand side).
- Read the measure of BD on the protractor. Since BD is on the left - hand side of the protractor, the measure is 180°−40° = 140° (because the 40° mark is on the inner scale? No, the outer scale: from 0° (BC) to 180° (BA), so BD is at 140° (180°−40°), so ∠DBC = 140°.

2. ∠DBE:
- Sides are BD and BE.
- We know ∠DBC = 140°, and ∠EBC: the angle between BE and BC. From the protractor, BE is at 70° from BC (because the mark for BE is 70° from BC on the outer scale? Wait, the outer scale (0° at BC, 180° at BA) at BE is 70°? No, the outer scale at BE is 110°? Wait, I think the correct measures are:

- ∠DBC = 140°
- ∠DBE = 70° (because the angle between BD and BE: BD is 40° from BA, BE is 110° from BA (180°−70°), so 110°−40° = 70°)
- ∠EBF = 30° (because BE is 110° from BA, BF is 140° from BA (180°−40°), so 140°−110° = 30°)
- ∠DBF = 100° (because BD is 40° from BA, BF is 140° from BA, so 140°−40° = 100°)

Wait, let's start over with the correct protractor reading:

- The straight line is AC, with B as the vertex.
- For ∠DBC: The angle between BD and BC. Since BC is along the positive x - axis (0°), and BD is at an angle of 140° (because the protractor shows that from BC (0°) to BD, the angle is 140° (using the outer scale, which goes from 0° at BC to 180° at BA)). So ∠DBC = 140°.
- For ∠DBE: The angle between BD and BE. The angle between BE and BC is 70° (from the outer scale, 0° at BC to BE is 70°? No, the outer scale at BE is 110°? Wait, I think the correct measures are:

After carefully looking at the protractor:

- ∠DBC: 140° (because BD is 40° from BA, so 180 - 40 = 140° from BC)
- ∠DBE: 70° (because BE is 110° from BC, so 140 - 70 = 70°)
- ∠EBF: 30° (because BF is 40° from BC, so 70 - 40 = 30°)
- ∠DBF: 100° (because 140 - 40 = 100° or 70+30 = 100°)

Step1: Measure ∠DBC

The angle between BD and BC. Since BC is a straight line (0°) and BD is at 140° from BC (using the protractor's outer scale: 180° - 40° = 140°), so $\angle DBC = 140^\circ$.

Step2: Measure ∠DBE

The angle between BD and BE. We know $\angle DBC = 140^\circ$ and the angle between BE and BC is $70^\circ$ (from the protractor). So $\angle DBE=\angle DBC - \angle EBC=140^\circ - 70^\circ = 70^\circ$.

Step3: Measure ∠EBF

The angle between BE and BF. The angle between BE and BC is $70^\circ$ and the angle between BF and BC is $40^\circ$ (from the protractor). So $\angle EBF=\angle EBC - \angle FBC = 70^\circ - 40^\circ = 30^\circ$.

Step4: Measure ∠DBF

The angle between BD and BF. We can calculate it as $\angle DBF=\angle DBC - \angle FBC=140^\circ - 40^\circ = 100^\circ$ (or $\angle DBF=\angle DBE+\angle EBF = 70^\circ+30^\circ = 100^\circ$).

Answer:

$\angle DBC = 140^\circ$, $\angle DBE = 70^\circ$, $\angle EBF = 30^\circ$, $\angle DBF = 100^\circ$