- #1

mrhingle

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- Thread starter mrhingle
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- #1

mrhingle

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- #2

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- #3

rl.bhat

Homework Helper

- 4,433

- 9

Show your calculations.

- #4

mrhingle

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[itеx] v(a)^2 = 9^2 +4.5^2 - 2 * 4.5 * 9 cos (50)

v(a) = 7.013

7.013/Sin(50) = 4.5/ Sin(alpha)

alpha = 29.44

7.013/sin(50) = 9/sin(beta)

beta = 79.44

[/itеx]

- #5

mrhingle

- 21

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[itеx] v(a)^2 = 9^2 +4.5^2 - 2 * 4.5 * 9 cos (50)

v(a) = 7.013

7.013/Sin(50) = 4.5/ Sin(alpha)

alpha = 29.44

7.013/sin(50) = 9/sin(beta)

beta = 79.44

[/itеx]

- #6

rl.bhat

Homework Helper

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- #7

mrhingle

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used the law of cosines. still got 29.44

- #8

rl.bhat

Homework Helper

- 4,433

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Sorry. Use cosine rule to find the angle between 4.5 and 7.013.

- #9

mrhingle

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79.44

- #10

rl.bhat

Homework Helper

- 4,433

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Is there no negative sign in the answer?

- #11

mrhingle

- 21

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nope

- #12

mrhingle

- 21

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the inverse is negative

- #13

Chestermiller

Mentor

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The angle beta could be either 79.44 or (180 - 79.44), as determined by the law of sines.

Last edited:

- #14

mrhingle

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what does that mean? I thought this was a law? Do you have to use the same law for all angles?

- #15

azizlwl

- 1,065

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Sides:

a = 7.01315 m

b = 9 m

c = 4.5 m

Angles:

A = 50°

B = 100.559°

C = 29.4415°

- #16

rl.bhat

Homework Helper

- 4,433

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what does that mean? I thought this was a law? Do you have to use the same law for all angles?

In the second quadrant cosine is negative. So whatever angle you get, you have to write it as 180-θ

- #17

mrhingle

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- #18

Chestermiller

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sin(θ) = sin(180 - θ)what does that mean? I thought this was a law? Do you have to use the same law for all angles?

- #19

azizlwl

- 1,065

- 10

It how the law was derived.

In this example one of the common perperdicular line(common to adjacent angles) is outside the triangle.

Thus we are measuring the external angle.

Thanks for bring it up. I think that it is just plug and chug.

Add: It's really prove that a diagram or a sketch is very helpful as in tool of problem solving, IDEA(D for drawing) from Richard Wolfson's book.

Last edited:

- #20

Chestermiller

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No the laws are correct. There are two possible solutions obtained applying the law of sines to this problem. Both solutions satisfy the law of sines. Applying the law of cosines to the other angles of the triangle clinches the results. In fact, applying the law of cosines delivers a unique solution.

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