# Finding the value of A and B (Another problem)

1. Nov 28, 2015

### shayaan_musta

Hello experts,
I have attached the figure below kindly see it. And here is my calculation,
Using law of sine,
from figure we can say that α+β=120°
and β=120°-α
and sin(β)/sin(α)=PA/PB=3/40=0.075

so,
sin(120°-α)=sin(120°)cos(α)-cos(120°)sin(α)
since, sin(β)=sin(120°-α)
so,
sin(β)=sin(120°)cos(α)-cos(120°)sin(α)
dividing both sides by sin(α)
so,
sin(β)/sin(α)=sin(120°)cos(α)/sin(α)-cos(120°)sin(α)/sin(α)
sin(β)/sin(α)=sin(120°)cos(α)/sin(α)-cos(120°)
sin(β)/sin(α)=0.866*cos(α)/sin(α)-(-0.5)
3/40=0.866*cot(α)+0.5
0.075=0.866*cot(α)+0.5
-0.425=0.866*cot(α)
cot(α)=-0.425/0.866
cot(α)=-0.49
α=-63.86°

What is this? Why α is -ve? Where I have gone wrong? From figure we can clearly see that it is more than 90°. But calculation says it is -63.86°.

According to the book point A is at -3.70 and B is at -53.35. But I am not getting my angle correct. HElp

2. Nov 28, 2015

### haruspex

What other solutions are there to cot(α)=-0.49?

3. Nov 28, 2015

### SteamKing

Staff Emeritus
Nope, I don't see any figure.

4. Nov 29, 2015

### shayaan_musta

Other solutions mean?

5. Nov 29, 2015

### shayaan_musta

Don't you see the figure of triangle?

#### Attached Files:

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6. Nov 29, 2015

### haruspex

There's more than one angle in a 360 degree range with a cotangent equal to a given value. For a positive value, there will be one in the first quadrant and one in the third. Here you have a negative value.

7. Nov 29, 2015

### shayaan_musta

No. I dont understand your point of view about more than one angle. Please elaborate with example.

8. Nov 29, 2015

### SteamKing

Staff Emeritus
I do now that you have posted the figure.

9. Nov 29, 2015

### SteamKing

Staff Emeritus
This figure shows the geometric representation of the cotangent:

http://www.regentsprep.org/regents/math/algtrig/att5/unitcircletriglabel.jpg
This particular figure shows the cotangent for an angle θ which is in the first quadrant. Like its cousin the tangent, cotangent is positive for angles in the first and third quadrants, but negative for angles in the second and fourth quadrants.

You have to figure out how your result relates to your original triangle.

10. Nov 29, 2015

### shayaan_musta

11. Nov 29, 2015

### SteamKing

Staff Emeritus
You have different triangles.

In the first problem, angle α was less than 90°. In the current problem, angle α is greater than 90°. Don't you think that makes a difference in the value of cot(α)?

12. Nov 29, 2015

### shayaan_musta

Yes, you are right. I have already observed it. But I my calculation is for α so why my α is not correct?
If 180-63.86=116.14

This is greater than 90. And from figure we can see that α is also greater than 90.

If 116.14 is the value for what I am looking for then I don't that understand that why this time I should subtract it from 180. why not in the last question I did this.

Much confusion :'(

13. Nov 29, 2015

### SteamKing

Staff Emeritus
I don't think anyone is suggesting that angle α be subtracted from 180°. It's just that your original value of α = -63.86°. What you must do is reconcile this result with the triangle which shows that α > 90°.

14. Nov 29, 2015

### shayaan_musta

No. I really don't understand.

I have calculated the value of α, then what should be further can I do with it?

And you are true no one suggested me to subtract it. But I don't see any further solution.

15. Nov 29, 2015

### SteamKing

Staff Emeritus
What if you added angle α algebraically to 180°? Wouldn't that give you a result which satisfies the triangle? When figuring angles, references are important.

16. Nov 29, 2015

### shayaan_musta

I don't want to lost in the maze. Kindly let me follow the light not to the darkness.

I am closing thread. I don't find any solution to my question.

17. Nov 29, 2015

### SteamKing

Staff Emeritus
That's too bad. You really should review trigonometry to get a better handle on it. Trig pops up in all kinds of math and engineering.

18. Nov 29, 2015

### shayaan_musta

I don't have time to review it. I have exam tomorrow on 30th November, 2015. 3 years ago we have trigonometry as a subject. After 3 years it comes again and in this 3 years we only have integration more than trigonometry.

19. Nov 29, 2015

### SteamKing

Staff Emeritus
Doesn't matter. Trig is one of those things which you use frequently. It doesn't matter if you studied it 3 years ago or 40 years ago like I did. You have to keep on top of it or you could risk blowing an exam.

20. Nov 29, 2015

### shayaan_musta

OK just solve me this problem to the value of α then further I will try on my own. But kindly find the exact value for me of α

21. Nov 29, 2015

### Samy_A

As in your previous exercise, draw the line from P perpendicular to the horizontal axis, and let Q be the point where it intersects the horizontal axis.

You already have $\cot(\alpha)=-0.49$.
As others have explained, you have two solutions for this, and you obviously need the one in the second quadrant.
So $\alpha=2.027$ (in radians, something like 116°).

In the right triangle APQ:
$|AQ|=\cot(\pi-\alpha)*|PQ|=0.49*3.464=1.7$
So the coordinates of A are $(-2-1.7,0)=(-3.7,0)$.

Use the triangle BPQ to get the coordinates of B.

22. Nov 29, 2015

### shayaan_musta

Hello Samy_A,

I was looking your way before my eyes. hehehehehe...

Thank you very much once again.

One problem still drowning me into the sea of confusion from the starting of this thread is that what does two solutions meant as everybody said and you also said that. I don't get it. Can you elaborate in simple way?

23. Nov 29, 2015

### Samy_A

There is not much to elaborate on.

The cotagent is negative in the second and fourth quadrant, so in your case $\cot(\alpha)=-0.49$ implied that $\alpha=2.027$ (close to d116°) or that $\alpha=-1.11$ (close to -63°). By inspection, only the first solution makes sense for you triangle.

Last edited: Nov 29, 2015
24. Nov 29, 2015

### shayaan_musta

In the previous problem we have α=88° and it is positive as it should be in second quadrant. That's why we have not subtracted it from 180° anymore.

But here we have α=-63.853° and it is negative as it should not be in the second quadrant. That's why we have this 180°-63.853°=116.147°(in radians, 2.027rad)

Am I getting it right @Samy_A?

25. Nov 29, 2015

### Samy_A

Both angles are mathematically correct solutions of the $\cot (\alpha)=-0.49$ equation. But only the second one makes sense in your specific situation.