Use the diagram for questions 2-4.
2. Suppose mBC-67.37°. Determine mAB and mLADB.
3. BE DE; BE-6; and AC-13. Find AE.
C
D
B
4. Using the measurements provided in Question 3, solve for BC. Round your answer to the nearest hundredths plac
or leave in simplest radical form.

Use the diagram for questions 242 Suppose mBC6737 Determine mAB and mLADB3 BE DE BE6 and AC13 Find AECDB4 Using the measurements provided in Question 3 solve fo class=


Answer :

Answer:

  2. arc AB = 112.63°; angle ADB = 56.315°

  3. AE = 9

  4. BC = 2√13 ≈ 7.21

Step-by-step explanation:

You want various arcs, angles, and segment measures in the circle diagram given.

2. Arc AB

Since AC is a diameter of the circle, arc AC will be 180°. Arc AB is the difference between that and arc BC:

  Arc AB = 180° -67.37°

  Arc AB = 112.63°

Angle ADB is an inscribed angle subtending arc AB, so the angle measure is half the arc measure:

  Angle ADB = 1/2(112.63°)

  Angle ADB = 56.315°

3. AE

The product of segments AE and EC is the same as the product of segments BE and ED:

  AE(13 -AE) = 6·6

  9·(13 -9) = 9·4 = 36

The length of segment AE is 9 units.

4. BC

The length of BC is the root of the product of CE and AC:

  BC = √(4·13)

  BC = 2√13 ≈ 7.21

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Additional comment

The quadratic equation in 3 can be solved a number of ways. We initially found the solution by graphing. It can also be found by considering factors of 36 that total 13. We know AE > CE, so AE is the greater of the two factors 4 and 9.

The relation used in 4 is a consequence of the similarity of triangles CBE, BAE, and CAB. This similarity means CE/CB = CB/CA, or CB² = CE·CA. You could also use the Pythagorean relation CB² = CE² +BE² to get the same result.

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