Cosine rule

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Homework Statement


ABC is a triangle
The angle A = 30 degrees
AB = 7cm
BC = 5 cm
C is an acute angle. Find the size of angle C?

Homework Equations


I need to use the cosine rule here Cos C = (a^2 + b^2 - c^2) / 2ac

The Attempt at a Solution


The problem is I do not know the size of side b?
So how can I attempt to resolve the answer?

Cos C = (5^2 + b^2 - 7^2) / 2*5*7
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


ABC is a triangle
The angle A = 30 degrees
AB = 7cm
BC = 5 cm
C is an acute angle. Find the size of angle C?

Homework Equations


I need to use the cosine rule here Cos C = (a^2 + b^2 - c^2) / 2ac

The Attempt at a Solution


The problem is I do not know the size of side b?
So how can I attempt to resolve the answer?

Cos C = (5^2 + b^2 - 7^2) / 2*5*7
Why do you need to use the cosine rule? Did somebody tell you that you are being forced to use it?

Just use whatever tool is the most convenient. If I were doing it I would let the points be at ##A = (0,0), B = (7,0), C = (x,y).## You can use the information about angle A and distance BC to get two equations in ##x, y##. These can be reduced to a single equation in ##x##, which is solvable without too much trouble.
 
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Thanks Ray but we have never used this method in class and I am sure this is the approach my teacher wants us to do it. Can I not use the cosine rule or sine rule then?
 
  • #4
gneill
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One way or another you'll have to find side b if you're going to apply the cosine rule. If you're not comfortable jumping right into the algebra with a method you've not used before, then start with a diagram and come up with a plan from there.

upload_2018-12-18_12-29-12.png
 

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  • #6
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Can I not use the cosine rule or sine rule then?
What about this second option ? :smile:
 
  • #7
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Ah ha!

cos A / a = sin C / c

So cos 30 / 5 = sin C / 7

So sin C = 7*(cos 30 / 5) = 1.212435565

How do I get the angle?
 
  • #8
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  • #9
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Ah ha!

sin A / a = sin C / c

So sin 30 / 5 = sin C / 7

So sin C = 7*(sin 30) / 5 = 0.7

arcsin gives 44.4 degrees
 
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In general you will use the cos rule for solving a triangle if the givens are the 3 sides or alternatively 2 sides and an angle formed between the 2 given sides.
 
  • #11
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Ah ha!

sin A / a = sin C / c

So sin 30 / 5 = sin C / 7

So sin C = 7*(sin 30) / 5 = 0.7

arcsin gives 44.4 degrees
It could be that there are two different triangles that satisfy the given measurements, as in the drawing in post #4. If ##\sin(C) = .7## there is also another angle that is possible for C.

Edit: Since it is stated in post #1 that C is an acute angle, the above does not apply.
 
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  • #13
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It could be that there are two different triangles that satisfy the given measurements, as in the drawing in post #4. If sin(C)=.7sin⁡(C)=.7\sin(C) = .7 there is also another angle that is possible for C.
Actually, that was my primary reason for posting the diagram. Too subtle?
 
  • #15
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Maybe too subtle, since the OP didn't seem to pick up on it.
:frown: Okay, will a tad more blatant in future. I hate blatant when gently nudging :smile:
 
  • #16
SammyS
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Actually, that was my primary reason for posting the diagram. Too subtle?
I thought that it was just subtle enough.
 
  • #17
symbolipoint
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I drew the diagram according to original problem description, and intuition tells me that b, and angle B, and angle C are not locked into any set of single values. Not a unique triangle. That's only first impression.

Thinking further,
forget about trying Law Of Cosines.

Law Of Sines seems fitting:

sin(C)/7=sin(30)/5

sin(C)=(7/5)sin(30)

sin(C)=7/10
and from this you could get value for angle C.


Now, you could use AB, BC, and angle at B, to find b, using Law Of Cosines.; but you could use Law Of Sines again if you want.
 
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I drew the diagram according to original problem description, ....
Why come with this now, when OP has cracked the case in #9 ?
 
  • #19
symbolipoint
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Why come with this now, when OP has cracked the case in #9 ?
Post was not old. I had to try the problem myself to see if or where Law Of Cosines would be useful; and I edited my own post twice within 20 minutes. Post #9 did not indicate any use of Law Of Cosines.
 
  • #20
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Thanks Ray but we have never used this method in class and I am sure this is the approach my teacher wants us to do it. Can I not use the cosine rule or sine rule then?
 

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