# Math - Trig algebra problem

1. Jul 11, 2009

### flyingpig

1. The problem statement, all variables and given/known data

T₁cosθ₁+T₂cosθ₂=mg
T₁sinθ₁=T₂sinθ₂

Part A Isolating T₁ terms on one side,

T₁cosθ₁=mg‐T₂cosθ₂
T₁sinθ₁=T₂sinθ₂

Part B Dividing the two equations, using tanθ=sinθ/cosθ,

cosθ₁/sinθ₁=mg/(T₂sinθ₂)‐cosθ₂/sinθ₂

1/tanθ₁=mg/(T₂sinθ₂)‐1/tanθ₂

2. Relevant equations

SOH CAH TOA

3. The attempt at a solution

I copy and paste the problem from my textbook. I will show you my attempt at this problem.
My problem lies in Part B.

It says that I should use tanθ=sinθ/cosθ, but it flips the equation.

I want to set T₂ as 18.
I want to set mg as 22.
I want to set θ₂ as 64 degrees

Accordingly tanθ=sinθ/cosθ, would then be (18sinθ₂)/(mg-18cosθ₂) which would result = 1.1466.

(18sin64)/(22-18cos64)=

But if I plug the numbers that I set above, it would give me 1.402

mg/(T₂sinθ₂)‐cosθ₂/sinθ₂ = 22/(18sin64) - cos64/sin64 = 1.402

2. Jul 11, 2009

### Coto

Hi flyingpig,

I would try typing this into your calculator again. For your very last step, I get 0.872 instead of 1.402 which if you take 1/0.872, you get 1.1466 as above.

Hope this helps.

3. Jul 11, 2009

### flyingpig

I got it, but can you tell me why and how it is flipped?

4. Jul 11, 2009

### Coto

In the first case, tanθ₁ = (T₂sinθ₂)/(mg-T₂cosθ₂). Plugging in your numbers yields 1.1466. That is:

$$\tan \theta _1 = 1.1466$$

In the second case (1/tanθ₁=mg/(T₂sinθ₂)‐1/tanθ₂), plugging in your numbers yields 0.872. That is:

$$\frac{1}{\tan \theta _1} = 0.872$$

So you can see that you've actually calculated 1/tan in the second case instead of tan like in the first case.

5. Jul 11, 2009

### flyingpig

Why do we even flip it?

6. Jul 11, 2009

### Coto

Perhaps you're a bit confused with what you've done.

Starting from your original post. In part B you have written some equations down. Here you took equation 1 from part A and divided it by equation 2 from part A to get a 1/tan on one side and the rest on the other. This is what was done in part B. Now it would just be a matter of plugging in your numbers to get a solution.

However, in your "Attempt at the solution" section, for some reason, you re-did part B, but this time you divided equation 2 from Part A by equation 1 from Part A and hence you got a tan instead of a 1/tan, hence for the two solutions to compare you would have to do like I did in post #4.

7. Jul 11, 2009

### flyingpig

cosθ₁/sinθ₁=mg/(T₂sinθ₂)‐cosθ₂/sinθ₂

mg/(T₂sinθ₂)‐cosθ₂/sinθ₂ <--- I want to focus on more on this part.

It is a problem from my textbook, I got confused by the bracket.

Originally the tan ratio is (T₂sinθ₂)/(mg - T₂cosθ₂), but when I flipped it, it should look like (mg - T₂cosθ₂)/(T₂sinθ₂).

(mg - T₂cosθ₂)/(T₂sinθ₂).
I know this is wrong, but how do I make the reciprocal of this and turn it into the verison my textbook had. Because I am really confused how my textbook seperate the "cosθ₂/sinθ₂" from the (mg - T₂cosθ₂)/(T₂sinθ₂) equation. Also, if you can don't use the math operators like I have.... that's probably where my question started. Thanks.

8. Jul 11, 2009

### Coto

Okay, well it seems like you're stuck on just some basic algebra:

$$\tan \theta _1 = \frac{T_2 \sin \theta _2}{mg - T_2 \cos \theta _2} \implies \frac{1}{\tan \theta _1} = \frac{mg - T_2 \cos \theta _2}{T_2 \sin \theta _2}$$

Now for the part you seem to be stuck on:

$$\frac{mg - T_2 \cos \theta _2}{T_2 \sin \theta _2} = \frac{mg}{T_2 \sin \theta _2} - \frac{T_2 \cos \theta _2}{T_2 \sin \theta _2} = \frac{mg}{T_2 \sin \theta _2} - \frac{1}{\tan \theta _2}$$

Let me know if you have any questions, but beyond this we are getting into some really basic algebra.

9. Jul 11, 2009

### flyingpig

You've cleared up ALOT! However....

$$\frac{mg - T_2 \cos \theta _2}{T_2 \sin \theta _2} = \frac{mg}{T_2 \sin \theta _2} - \frac{T_2 \cos \theta _2}{T_2 \sin \theta _2}$$

This is troubling me

$$\frac{T_2 \cos \theta _2}{T_2 \sin \theta _2}$$

I am not saying you are incorrect, because I believe you are correct. But the textbook canceled out the $$T_2$$s

If they do they would it still make it equal? I am sorry for wasting your time on such a simple matter...I would've probably attracted more attention if I knew how to do nice operators.

10. Jul 11, 2009

### Coto

I neglected canceling out the $$T_2$$s just to show you the step where I split the fraction. You could have done the cancellation then too.

Notice in the last step, I make the cancellation, and the final answer is the same.

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