How is the Sine law written for this problem?

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In summary, the resultant vector of vectors A and B, shown in the figure, can be found using the cosine law and sine law. The resultant vector has a magnitude of 41.39 m/s and makes an angle of 72.70 degrees clockwise from the positive x-axis. The sine law equation used to calculate the angle is ##\large \frac {R}{sin70_{0}} = \frac{17}{sin \alpha}##, where 70 degrees is the full angle and 17 is the length opposite to the angle alpha. There is also an alternative method of using the tangent rule to calculate alpha.
  • #1
Benjamin_harsh
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Homework Statement
Find the resultant vector of vectors A and B shown in the figure.
Relevant Equations
##\large \frac {R}{Sin 70^0} = \frac {17}{Sin\alpha}##
Find the resultant vector of vectors A and B shown in the figure.

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Solution:
By geometry method:

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Cosine law for the right side triangle.

##R^{2} = 17^{2} + 44^{2} - 2 (17)(14).cos 70^{0}##

##R = 41.39 m/sec##

By Sin law,

##\large\frac {R}{Sin 70^0} = \frac {17}{Sin\alpha}##

##sin \alpha = \large \frac{17.sin 70^{0}}{R} = \frac{17.sin 70^{0}}{41.39}##

##\alpha = 22.70^{0}##

##\theta_{x} = 50 + \alpha = 50 + 22.70##

##\theta_{x} = 72.70^{0}##
 
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  • #2
Is there a question here ?

PS tip: in ##\LaTeX##, use \sin and not Sin or sin
Idem \cos, \tan, \log, etc.
 
  • #3
Benjamin_harsh said:
By Sin law,

##\large\frac {R}{Sin 70^0} = \frac {17}{Sin\alpha}##

##sin \alpha = \large \frac{17.sin 70^{0}}{R} = \frac{17.sin 70^{0}}{41.39}##
Expanding a bit on @BvU's comments regarding ##\LaTeX :##

Your code for the expression:
##\large\frac {R}{Sin 70^0} = \frac {17}{Sin\alpha}##

is: ##\large\frac {R}{Sin 70^0} = \frac {17}{Sin\alpha}##.

As BvU recommends, use "\sin" rather than "Sin".
Also, there is no degree symbol built into ##\LaTeX ##. Using "^0", as you did, is one work-around. Another is to use "^\circ".
You can get large fractions by using "\dfrac" rather than "\large\frac". Using "\large" the way you did makes everything that follows large, not just the fraction. You could avoid making all that large by 'wrapping' the expression in braces: " { \large \frac{num}{denom} } ".

Using the following code: ##{\large\frac {R}{\sin 70^\circ}} = \dfrac {17}{\sin\alpha}## gives the result:

##{\large\frac {R}{\sin 70^\circ}} = \dfrac {17}{\sin\alpha}##
(The first fraction is done using the "\large" feature)

One more thing:
Use "\cdot" rather than a period, ".", for multiplication.

##\sin \alpha = \dfrac{17 \cdot \sin 70 ^\circ }{R} = \dfrac{17 \cdot \sin 70^\circ }{41.39}## gives the result:

##\sin \alpha = \dfrac{17 \cdot \sin 70 ^\circ }{R} = \dfrac{17 \cdot \sin 70^\circ }{41.39}##
 
  • #4
I am not asking about LATEX. I am asking about concept.
 
  • #5
Benjamin_harsh said:
I am not asking about LATEX. I am asking about concept.
Yup. Looks good.
So your resultant vector has a magnitude of 41.39 m/s and makes an angle of ##72.70^\circ## clockwise from the positive ##x##-axis.​

You can check your answer using x-components and y-components.

The ##\LaTeX## tips were given free of charge by @BvU and me. :smile:
 
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  • #6
Please explain this sin law equation. ##\large \frac {R}{sin70_{0}} = \frac{17}{sin \alpha}##.?
##70^{0}## is not the full angle, It is just a portion of the angle. So how sine law is calculated on portion of the angle?
 
  • #7
It's the full angle: the 70##^\circ## is shown in the right half of the figure (that you should redraw a bit more to scale).
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  • #8
How ##17## included in the sine law? See, in the diagram, ##17## is far away from ##\alpha## so how it is included ?
 
  • #9
Did you already re-draw the picture a bit more to scale as I suggested ?
Then you'll see more easily where the 17 is sitting: opposite the angle ##\alpha##.
Please do a little bit of pondering answers before shooting off yet another question :wink:

244732
 
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  • #11
Demonstrating a mstery of which drawing program, exactly ?
 
  • #12
BvU said:
Demonstrating a mastery of which drawing program, exactly ?
I used LibreOffice Draw.

To call my abilities to use it "mastery" would be a stretch. I almost gave up using it (for making this figure) but then made a small break-through with making circular arcs. I can now make the arc I want in about half of my attempts.
 
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  • #13
Benjamin_harsh said:
Please explain this sin law equation. ##\large \frac {R}{sin70_{0}} = \frac{17}{sin \alpha}##.?
##70^{0}## is not the full angle, It is just a portion of the angle. So how sine law is calculated on portion of the angle?
For the record, it's also possible to calculate ##\alpha## without recourse to R or the sin rule. Referring to SammyS's very excellent diagram, drop a perpendicular from where the green arrows meet to the side of length 44. Then: $$ \tan{\alpha} = \frac{17\sin{70^\circ}}{44-17\cos{70^\circ}} $$ This is the 'tan rule' but unfortunately nobody ever teaches it in school!

PS: have picked up all the above 'tips' re use of ## \LaTeX ##!
 
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Related to How is the Sine law written for this problem?

1. What is the Sine law?

The Sine law, also known as the Law of Sines, is a mathematical relationship between the angles and sides of a triangle. It states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is constant for all the sides and angles of the triangle.

2. How is the Sine law written?

The Sine law is typically written in the form of a proportion: sin(A)/a = sin(B)/b = sin(C)/c, where A, B, and C are the angles of the triangle and a, b, and c are the lengths of the corresponding sides.

3. What does the Sine law help us solve?

The Sine law is used to solve for missing angles or sides in a triangle when we know the measurements of at least two angles and one side, or two sides and one angle.

4. Is the Sine law only applicable to right triangles?

No, the Sine law can be used for any type of triangle, whether it is acute, obtuse, or right angled.

5. Can the Sine law be used for non-triangular shapes?

No, the Sine law is only applicable to triangles and cannot be used for other shapes.

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