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lisov135 [29]
3 years ago
5

In ΔDEF, d = 6.4 inches, e = 7.8 inches and ∠F=33°. Find the length of f, to the nearest 10th of an inch.

Mathematics
1 answer:
Vitek1552 [10]3 years ago
5 0

9514 1404 393

Answer:

  4.3 in

Step-by-step explanation:

The Law of Cosines is applicable when you have two sides and the angle between them. The third side satisfies ...

  f² = d² + e² -2de·cos(F)

  f² = 6.4² + 7.8² -2·6.4·7.8·cos(33°) ≈ 18.0671

  f ≈ √18.0671 ≈ 4.2505

The length of f is about 4.3 inches.

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For this case we must solve the following quadratic equation:

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x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

Substituting:

x = \frac {-2 \pm \sqrt {2 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-2 \pm \sqrt {4 + 24}} {2}\\x = \frac {-2 \pm \sqrt {28}} {2}\\x = \frac {-2 \pm \sqrt {2 ^ 2 * 7}} {2}\\x = \frac {-2 \pm2 \sqrt {7}} {2}\\x = -1 \pm \sqrt {7}

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x_ {1} = - 1+ \sqrt {7}\\x_ {2} = - 1- \sqrt {7}

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Answer:

The answer is given below

Step-by-step explanation:

From the diagram below,Let the line AB and CD be parallel line. Let transversal line EF cut AB at Y and transversal line EF cut CD at Z.

The bisector of ∠BYZ and ∠DZY meet at O. Therefore ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2

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∠BYZ + ∠DZY = 180 (sum of consecutive interior angles)

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∠BYZ/2 + ∠DZY/2 = 90°

In ΔOYZ:

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But  ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2

∠DZY/2 + ∠BYZ/2 + ∠YOZ = 180

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∠YOZ = 180 - 90

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