# How to solve this Cosine Law Equation?

Question is regarding Scalars and Vectors Article.
Q: One of the two forces is double the other and their resultant is equal to the greater force . The angle between them is ?
Ans : Its answer is cos^-1 (-1/4)

My solution :
The formula for Cosine law is
R= √A²+B² +2ABcos theta

Our teacher told that when talking about force we can write :
F= √F1² + F2²+F1²X F2² cos theta
F= √F1+F2+2F²cos theta
F=√1+cos theta

But, what I did to solve problem is this below :-

1: I plugged in values . A = F1 , B = 2F , Resultant "R" = 2f

2: I got this equation :
= 2F = √F1² + 2F² +2F(2F) cos theta
= 2F= √F²(cancel) + 2F²(cancel) + 4F² cos theta : F² F² cancel each other
= 2F = √2(cancel)+4(cancel)F² cos theta : 2 and 4 are divided by 2 and cos theta shifts
= F/2F²= √ 1+ cos theta
= 1/4= √1 + cos theta
= cos ^-1 (-1/4)

But I know the way solved is not it is meant to be solved because I am in Pre- Medical and i have less knowledge of mathematics , so Kindly tell me how to deal with such numericals and please solve it and tell me shortest route to solve it . thanks in advance

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Mark44
Mentor
Question is regarding Scalars and Vectors Article.
Q: One of the two forces is double the other and their resultant is equal to the greater force . The angle between them is ?
Ans : Its answer is cos^-1 (-1/4)
This answer doesn't make sense. If the smaller force is F and the larger force is 2F, then we have the vector equation 2F + F = 2F, so 3F = 2F.
If the resultant of two vectors is equal to one of the vectors, that says something about the other vector.
Shahab Mirza said:
My solution :
The formula for Cosine law is
R= √A²+B² +2ABcos theta
Your sign of the 2ab cos(θ) term is wrong. The Law of Cosines is ##c^2 = a^2 + b^2 - 2ab cos(\theta)##, where θ is the angle between the sides a and b.

Furthermore, what triangle are you working with? The two vectors and their resultant don't form a triangle.
Shahab Mirza said:
Our teacher told that when talking about force we can write :
F= √F1² + F2²+F1²X F2² cos theta
The above isn't right.
Shahab Mirza said:
F= √F1+F2+2F²cos theta
F=√1+cos theta

But, what I did to solve problem is this below :-

1: I plugged in values . A = F1 , B = 2F , Resultant "R" = 2f

2: I got this equation :
= 2F = √F1² + 2F² +2F(2F) cos theta
= 2F= √F²(cancel) + 2F²(cancel) + 4F² cos theta : F² F² cancel each other
?? Why do you think they cancel?
Shahab Mirza said:
= 2F = √2(cancel)+4(cancel)F² cos theta : 2 and 4 are divided by 2 and cos theta shifts
= F/2F²= √ 1+ cos theta
= 1/4= √1 + cos theta
= cos ^-1 (-1/4)

But I know the way solved is not it is meant to be solved because I am in Pre- Medical and i have less knowledge of mathematics , so Kindly tell me how to deal with such numericals and please solve it and tell me shortest route to solve it . thanks in advance

Last edited:
This answer doesn't make sense. If the smaller force is F and the larger force is 2F, then we have the vector equation 2F + F = 2F, so 3F = 2F.
If the resultant of two vectors is equal to one of the vectors, that says something about the other vector.
Your sign of the 2ab cos(θ) term is wrong. The Law of Cosines is ##c^2 = a^2 + b^2 - 2ab cos(\theta)##, where θ is the angle between the sides a and b.

Furthermore, what triangle are you working with? The two vectors and their resultant don't form a triangle.
The above isn't right.
?? Why do you think they cancel?

ok see this other solution sir , this is what I did now but in a simpler way.

2F² = (F)² + (2F)² + (2xF) (2F) cos theta
2F² = F² + 4F² + 4F² cos theta
2F² = F² + 8F² cos theta : numbers added 4F² + 4F² = 8F² and F² cancelled with F² of (8F²)
2F² = (1 + cos ) 8
2F²/8 = 1 + cos theta)
1/4
so cos inverse ^-1 (-1/4) answer

.
Furthermore, what triangle are you working with? The two vectors and their resultant don't form a triangle.
Yes exactly such numericals appear in our tests without any thing about triangle , I know law of cosine is obviously to find angle of triangle but I dont know why they are using law of cosine here?

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Mark44
Mentor
ok see this other solution sir , this is what I did now but in a simpler way.
Simpler, yes, but still wrong.
Shahab Mirza said:
2F² = (F)² + (2F)² + (2xF) (2F) cos theta
• The sign (marked in red) is wrong.
• On the left side, it should be (2F)2.
Shahab Mirza said:
2F² = F² + 4F² + 4F² cos theta
2F² = F² + 8F² cos theta : numbers added 4F² + 4F² = 8F² and F² cancelled with F² of (8F²)
No, you can't do this. 4F2 + 4F2cos(θ) ≠ 8F2cos(θ).
That's like saying that 4 + 4*7 = 8*7, which isn't true.
Shahab Mirza said:
2F² = (1 + cos ) 8
2F²/8 = 1 + cos theta)
1/4
so cos inverse ^-1 (-1/4) answer

I have actually used plus sign because If I would have used negetive sign than in answer sign would have been automatically reversed if cos >120 degree i think .

Btw please tell me how to solve it now ?

vela
Staff Emeritus
Homework Helper
You're on the right track. You just need to pay more attention to the details because you're making careless mistakes. Mark already pointed out various errors you made. Fix those and try again.

Simpler, yes, but still wrong.

• The sign (marked in red) is wrong.
• On the left side, it should be (2F)2.
No, you can't do this. 4F2 + 4F2cos(θ) ≠ 8F2cos(θ).
That's like saying that 4 + 4*7 = 8*7, which isn't true.

Ok here is what I did more bro
You're on the right track. You just need to pay more attention to the details because you're making careless mistakes. Mark already pointed out various errors you made. Fix those and try again.

Thanks alot , I found that I was doing careless simpler mathematical mistakes .
So I found this the correct solution .

= 2F = √F²+(2F)² + 2(F) (2F) cos theta
= (2F) = (√F²+(2F)² + 2(F) (2F) cos theta )² : squaring both sides
= 4F² = F² + 4F² + 4F² cos theta
= 4F² = 5F² + 4F² cos theta
= -5F² + 4F² =4F² cos theta
= -F² = 4F² cos theta
= -F² / 4F² = cos theta : F² cancelled with F²
= -1 / 4 = cos theta
= -1 / 4 / cos theta
= cos inverse ^ -1 (-1/4) . Final Answer

Mark44
Mentor
Shahab Mirza said:
So I found this the correct solution .

= 2F = √F²+(2F)² + 2(F) (2F) cos theta
= (2F) = (√F²+(2F)² + 2(F) (2F) cos theta )² : squaring both sides
= 4F² = F² + 4F² + 4F² cos theta
= 4F² = 5F² + 4F² cos theta
= -5F² + 4F² =4F² cos theta
= -F² = 4F² cos theta
= -F² / 4F² = cos theta : F² cancelled with F²
= -1 / 4 = cos theta
= -1 / 4 / cos theta
= cos inverse ^ -1 (-1/4) . Final Answer

In post #1 you gave this as the Cosine Law:
R= √A²+B² +2ABcos theta
This formulation is applicable if the angle ##\theta## is greater than 90° (see http://en.wikipedia.org/wiki/Law_of_cosines, in the discussion of Figure 2). What made you think that the angle between the two vectors was larger than 90°? Did you draw a sketch of the possible orientations of the two vectors?

There are several things wrong in your work above.
1. Each line in your work starts with '='. Except for the last two lines, each other line is an equation that already has '=' in it. The first 8 lines should NOT start with =.
2. The right side of the first equation above is √F²+(2F)² + 2(F) (2F) cos theta. What you wrote isn't what you meant, which was that the square root is of the whole expression, not just of F or F2. Use parentheses to indicate what you're taking the square root of, as in √(F²+(2F)² + 2(F) (2F) cos theta).
3. In the second line, where you indicate that you are squaring both sides, you show the left side as (2F), but this should be (2F)2. You don't "square an equation" -- you square each side of an equation.
4. In the 8th and 9th lines you have
= -1 / 4 = cos theta <--- this is an equation
5. = -1 / 4 / cos theta <--- this is NOT an equation
6. What do you mean by -1/4/cos theta? Are you dividing -1/4 by cos(theta)?
7. In the last line you have cos inverse ^ -1 (-1/4). Ignoring the fact that you have cos inverse AND an exponent that indicates the inverse of the cosine, what happened to theta? The last line should look like "theta = <some number>"

vela
Staff Emeritus
Homework Helper
I have actually used plus sign because If I would have used negetive sign than in answer sign would have been automatically reversed if cos >120 degree i think.
I think you meant 90 degrees, not 120 degrees. Anyway, that's not a good reason for changing the sign of that term, nor is it a valid reason.

Draw a sketch and make sure the ##\theta## in the equation actually represents what you think it does.

I think you meant 90 degrees, not 120 degrees. Anyway, that's not a good reason for changing the sign of that term, nor is it a valid reason.

Draw a sketch and make sure the ##\theta## in the equation actually represents what you think it does.
but at 90 degree angle force doesn't work , so I don't think this is meant to be used , I found this formulae in our textbook , which is used in all local examinations here , so if I will play with this formula then answer will not be desired , Thanks alot

In post #1 you gave this as the Cosine Law:
This formulation is applicable if the angle ##\theta## is greater than 90° (see http://en.wikipedia.org/wiki/Law_of_cosines, in the discussion of Figure 2). What made you think that the angle between the two vectors was larger than 90°? Did you draw a sketch of the possible orientations of the two vectors?

There are several things wrong in your work above.
1. Each line in your work starts with '='. Except for the last two lines, each other line is an equation that already has '=' in it. The first 8 lines should NOT start with =.
2. The right side of the first equation above is √F²+(2F)² + 2(F) (2F) cos theta. What you wrote isn't what you meant, which was that the square root is of the whole expression, not just of F or F2. Use parentheses to indicate what you're taking the square root of, as in √(F²+(2F)² + 2(F) (2F) cos theta).
3. In the second line, where you indicate that you are squaring both sides, you show the left side as (2F), but this should be (2F)2. You don't "square an equation" -- you square each side of an equation.
4. In the 8th and 9th lines you have
= -1 / 4 = cos theta <--- this is an equation
5. = -1 / 4 / cos theta <--- this is NOT an equation
6. What do you mean by -1/4/cos theta? Are you dividing -1/4 by cos(theta)?
7. In the last line you have cos inverse ^ -1 (-1/4). Ignoring the fact that you have cos inverse AND an exponent that indicates the inverse of the cosine, what happened to theta? The last line should look like "theta = <some number>"
Oh , actually I solved it in a manner we do in our exam mcqs where we have 1 minute to solve 1 question , let men explain you Sir.

This was my work
= 2F = √F²+(2F)² + 2(F) (2F) cos theta
= (2F) = (√F²+(2F)² + 2(F) (2F) cos theta )² : squaring both sides
= 4F² = F² + 4F² + 4F² cos theta
= 4F² = 5F² + 4F² cos theta
= -5F² + 4F² =4F² cos theta
= -F² = 4F² cos theta
= -F² / 4F² = cos theta : F² cancelled with F²
= -1 / 4 = cos theta
= -1 / 4 / cos theta
= cos inverse ^ -1 (-1/4) . Final Answer

so yes it was my mistake that I didnt wrote ² on (2F) so it is (2F)² = 4F²
and = yes it was wrong to mention it

and yes i divided -1/4 with cos theta to get cos inverse by shifting it . That was taught by teacher in our class . because answer isn't further solved , questions are weird in our local examinations they ask weird question which have weird solutions . Thanks for reply

but at 90 degree angle force doesn't work , so I don't think this is meant to be used , I found this formulae in our textbook , which is used in all local examinations here , so if I will play with this formula then answer will not be desired , Thanks alot

Sir , angle wasn't mentioned in question because these are quick quesitons which come in exam and at that time one is practically not in position to calculate in that way so they have tweaked cosin law for us and taught us to apply in that way , reason they haven' told because textbooks here are written by team to which it is difficult to speak or communicate their people

Mark44
Mentor
Sir , angle wasn't mentioned in question
This is what you have in post #1.
The angle between them is ?
So the bottom line should be something like "θ = cos-1(-1/4)"
Shahab Mirza said:
because these are quick quesitons which come in exam and at that time one is practically not in position to calculate in that way so they have tweaked cosin law for us and taught us to apply in that way , reason they haven' told because textbooks here are written by team to which it is difficult to speak or communicate their people

This is what you have in post #1.
So the bottom line should be something like "θ = cos-1(-1/4)"

ops sorry by angle I mean Triangle diagram or something like that wasn't mentioned in the question Sir .

Yes exactly this is the answer you are right .

I also have an quick alternate to solve this question

I found this method more quicker and applicable to variety of questions . See below I have made F understoodable in both 1F and 2F . So below is the less reliable but faster method .

simple use numbers without using "F"
1: 2 = √ 1² +2² + 2 X 2 cos theta
2: 2² =(√1² + 2² + 4 cos theta)² " square cancelled with underroot and 2 on left side squared to 4 .
2 : 4 = 1² + 4 + 4 cos theta
3: 4 = 5 + 4cos theta
4: -5+4 = 4 cos theta
5: -1 = 4 cos theta
6: -1/4 = cos theta

Thanks

Mark44
Mentor
That looks a lot better

That looks a lot better

Sorry for late reply sir , actually internet was down since days .

Yes this looks way better and way easy too because we have atleast 20 chapters of Physics with too much number of articles , so we have to find such a way that keeps thing in our mind for long term and save time , Thanks