Unit 1 geometry basics quiz 1-3 angle measures and relationships

1. From the diagram, it can be seen that side CD and side DG meet at a point D, resulting in them forming ∠4. Invariably, the sides of ∠4 are  side CD and side DG.

2. Vertex can be defined as the endpoint at which two sides meet to form an angle. From the diagram, the vertex of ∠2 then is  D.

3. ∠3 is an angle joined by line ED and line DG. Therefore, we can call ∠3, ∠EDG

4. Looking at ∠5 we can tell that it is less than 90°. All angles that are less than 90 are classified as acute angles, and thus, ∠5 can be classified as an acute angle.

5. Taking a deep look at ∠CDE. We can say that the angle is less than 180° even if its by a tiny fraction, and at the same time greater than 90°. Therefore, ∠CDE is said to be an obtuse angle.

6. If ∠5 is 42 and ∠1 is 117, then ∠CDF is an addition of both ∠5 and ∠1. Therefore, ∠CDF = 117 + 42 = 159°  

7. If ∠3 = 73°, then ∠FDE = ∠FDG - ∠3

∠FDE = 90 - 73

∠FDE = 17°

To learn more about angles in a triangle, see link

Answer/Step-by-step explanation:

1. Side CD and side DG meet at endpoint D to form <4. Therefore, the sides of <4 are:

Side CD and side DG.

2. Vertex of <2 is the endpoint at which two sides meet to form <2.

Vertex of <2 is D.

3. Another name for <3 is <EDG

4. <5 is less than 90°. Therefore, <5 can be classified as an acute angle.

5. <CDE is less than 180° but greater than 90°. Therefore, <CDE is classified as an obtuse angle.

6. m<5 = 42°

m<1 = 117°

m<CDF = ?

m<5 + m<1 = m<CDF (angle addition postulate)

42° + 117° = m<CDF (Substitution)

159° = m<CDF

m<CDF = 159°

7. m<3 = 73°

m<FDE = ?

m<FDG = right angle = 90°

m<3 + m<FDE = m<FDG (Angle addition postulate)

73° + m<FDE = 90° (Substitution)

73° + m<FDE - 73° = 90° - 73°

m<FDE = 17°

no ,sorry if you needed it but you can ask and try to see if any one will anwser ,

step-by-step explanation:

by good luck mark me as brainliest

have a nice day

step-by-step explanation: