# Can you me with this (it is all about vectors )

• donniemsb_12
In summary, the professor discussed an example in class involving a triangle with sides A = 2.0m at an angle of 40 degrees and B = 4.0m at an angle of 127 degrees. Using the law of cosines, she solved for the resultant R and the angle θ, which was approximately 62.2 degrees. The direction of R was described as 11.2 west of north, which was determined by adding the angle of R with the angle of A. The formula used, R^2 = A^2 + B^2 - 2AB cos(degree), is a constant for any triangle.
donniemsb_12
this example is being discussed by our professor in our class ! ( the one that i upload is the illustration made by our prof in our class )

this are the given and the solution :

A=2.0m @ 40'
B=4.0m @127'

SOLUTION :

R^2=A^2+B^2– 2AB Cos93'
R^2=(2.0)^2 + (4.0)^2 -(2)(2)(4)cos93'
R=4.56m

sin⁡θ/4=sin93'/4.56 sin⁡θ=62.2'

11.2 west of north

guys ! please explain to me how did they get the sin⁡θ=62.2' and 11.2 west of north and the formula ! thanks ! if u answer this i can now make the assignment given by our prof thanks a lot !

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What part don't you understand? Did you solve for sinθ? (That diagram is inaccurate, by the way.)

It's not sinθ = 62, but θ = 62. (approximately)

Doc Al said:
What part don't you understand? Did you solve for sinθ? (That diagram is inaccurate, by the way.)

It's not sinθ = 62, but θ = 62. (approximately)

that was being answered by our prof. that problem set as an example in our class ! ok please explain to me how did she get the answer of θ = 62' and the formula's ? and the " 11.2 west of north " where did it came from ? please help me thanks !

donniemsb_12 said:
i don't know ! that was being answered by our prof. that problem set as an example in our class ! ok please explain to me how did she get the answer of θ = 62' and the formula's ? and the " 11.2 west of north " please help me thanks !
Look up the http://en.wikipedia.org/wiki/Law_of_sines" . That's how she got the formula.

And once you've solve for the angle in the triangle to be 61.2', then that's 11.2' past vertical, which is 11.2 west of north.

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Doc Al said:
Look up the http://en.wikipedia.org/wiki/Law_of_sines" . That's how she got the formula.

And once you've solve for the angle in the triangle to be 61.2', then that's 11.2' past vertical, which is 11.2 west of north.

thanks sir ! now i know how did she get the 61.2 ! ! the thing that i don't know is the 11.2 west of north. please explain it more thanks !

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donniemsb_12 said:
thanks sir ! now i know how did she get the 61.2 ! ! the thing that i don't know is the 11.2 west of north. please explain it more thanks !
In the diagram, the resultant R makes an angle of 61.2 degrees with the vector A. Since vector A is at an angle of 40 degrees with the x-axis, that makes the resultant have an angle of 40 + 61.2 = 101.2 degrees. Which is 11.2 to the left of vertical (90 degrees). Which you can also describe as 11.2 west of north, if you take north as the +y axis.

Doc Al said:
In the diagram, the resultant R makes an angle of 61.2 degrees with the vector A. Since vector A is at an angle of 40 degrees with the x-axis, that makes the resultant have an angle of 40 + 61.2 = 101.2 degrees. Which is 11.2 to the left of vertical (90 degrees). Which you can also describe as 11.2 west of north, if you take north as the +y axis.

thanks ! i understand now ! that was a big help ! thank you so much ! can i ask more ? R^2=A^2+B^2– 2AB Cos(degree)' that formula ? is it always constant or it can be change base on the problem ?

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donniemsb_12 said:
thanks ! i understand now ! that was a big help ! thank you so much ! can i ask more ? R^2=A^2+B^2– 2AB Cos(degree)' that formula ? is it always constant or it can be change base on the problem ?
That's the http://en.wikipedia.org/wiki/Law_of_cosines" . It's a property of any triangle.

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Doc Al said:
That's the http://en.wikipedia.org/wiki/Law_of_cosines" . It's a property of any triangle.

thanks sir ! last question sir ! in the 11.2 west of north ! how did the direction get ? the "WEST OF NORTH " ?

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donniemsb_12 said:
thanks sir ! last question sir ! in the 11.2 west of north ! how did the direction get ? the "WEST OF NORTH " ?
In the diagram, up is north and left is west. (So north is 90' from the x-axis.)

Doc Al said:
In the diagram, up is north and left is west. (So north is 90' from the x-axis.)

thanks sir ! ! ! now i understand this ! thank you very much ! god bless !

## 1. What are vectors and how are they used in science?

Vectors are mathematical quantities that have both magnitude (size) and direction. They are commonly used in science to represent physical quantities such as velocity, force, and acceleration.

## 2. How do I add or subtract vectors?

To add or subtract vectors, you must first find their components (magnitude and direction) and then use the appropriate mathematical operations. For example, to add two vectors, you add their components separately in the same direction. To subtract, you subtract the components of the second vector from the first vector.

## 3. Can you explain the difference between position and displacement vectors?

A position vector represents the location of an object in a coordinate system, while a displacement vector represents the change in position of an object from its starting point to its ending point.

## 4. How do vectors relate to other scientific concepts, such as forces and motion?

Vectors are commonly used to represent forces and motion in science. For example, a force vector represents the direction and strength of a force, while a velocity vector represents the speed and direction of an object's motion.

## 5. Can you provide real-world examples of vector quantities?

Some common real-world examples of vector quantities include wind velocity, gravitational force, and the displacement of an object from its starting point to its ending point. Vectors are used to accurately represent and analyze these physical quantities in the field of science.

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