Finding the Angles: A Trigonometric Problem

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In summary, the conversation discussed finding all the angles from 0^{\circ} to 360^{\circ} inclusive which satisfy the equation \tan(x-30^{\circ}) - \tan 50^{\circ} = 0. The conversation involved using the trigonometric identity \tan(\alpha + \beta) = \frac{tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)} to solve the equation and the concept of periodicity in trigonometric functions to find all the solutions.
  • #1
omicron
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Find all the angles from [tex]0^{\circ}[/tex] to [tex]360^{\circ}[/tex] inclusive which satisfy the equation
[tex]$ \tan(x-30^{\circ}) - \tan 50^{\circ} = 0 [/tex]
 
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  • #2
If you want help, you need to show us what you've done so far, or what your thoughts are on how to go about solving it.
 
  • #3
I haven't done anything. I don't have a clue what to do.
 
  • #4
Here's a hint: write tan (x - 30) in terms of tan (50). What can you see then?
 
  • #5
[tex]\tan(\alpha + \beta) = \frac{tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}[/tex]

[tex]\tan(x - 30) = \frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}[/tex]

[tex]\frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}=\tan(50)[/tex]

[tex]\tan(x)+\tan(-30)=\tan(50) - \tan(x)\tan(-30)\tan(50)[/tex]

[tex]\tan(x) + \tan(x) \tan(-30)\tan(50)=\tan(50) - \tan(-30)[/tex]

[tex]\tan(x)(1 +\tan(-30)\tan(50))=\tan(50) - \tan(-30)[/tex]

[tex]\tan(x)=\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}[/tex]

:devil: :biggrin:
 
Last edited:
  • #6
Don't know :rolleyes:
 
  • #7
[tex]\tan(x-30^{\circ}) = \tan 50^{\circ}[/tex]

Can you go from there?
 
  • #8
Kahsi said:
[tex]\tan(\alpha + \beta) = \frac{tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}[/tex]

[tex]\tan(x - 30) = \frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}[/tex]

[tex]\frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}=\tan(50)[/tex]

[tex]\tan(x)+\tan(-30)=\tan(50) - \tan(x)\tan(-30)\tan(50)[/tex]

[tex]\tan(x) + \tan(x) \tan(-30)\tan(50)=\tan(50) - \tan(-30)[/tex]

[tex]\tan(x)(1 +\tan(-30)\tan(50))=\tan(50) - \tan(-30)[/tex]

[tex]\tan(x)=\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}[/tex]

:devil: :biggrin:

How did you get all that?? :confused:
 
  • #9
Kahsi said:
[tex]\tan(\alpha + \beta) = \frac{tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}[/tex]

[tex]\tan(x - 30) = \frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}[/tex]

[tex]\frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}=\tan(50)[/tex]

[tex]\tan(x)+\tan(-30)=\tan(50) - \tan(x)\tan(-30)\tan(50)[/tex]

[tex]\tan(x) + \tan(x) \tan(-30)\tan(50)=\tan(50) - \tan(-30)[/tex]

[tex]\tan(x)(1 +\tan(-30)\tan(50))=\tan(50) - \tan(-30)[/tex]

[tex]\tan(x)=\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}[/tex]

:devil: :biggrin:



:smile: It's not that complex:

[tex]\tan (x - 30) = \tan 50[/tex]

Hence to work out an initial value just apply arctan on both sides to get:

[tex]x - 30 = 50[/tex]
 
  • #10
There are two solutions to the problem, that is one of them.
 
  • #11
since the Tan curve goes in a period of 180 degrees, you take the value that you got as one of the solutions and add or subtract 180 to/from it, and every time the result is within the rang of 0 -360, so:

you do
[tex]\tan (x - 30) = \tan 50[/tex]
[tex]x = 80[/tex]

then

[tex]80 \pm 180n = x [/tex]

and the only other value that fits into the range is when

[tex]n = 1 [/tex]
[tex]80 + 180 = 260 [/tex]

Therefor the 2 answers are

[tex]x = 80, 260 [/tex]
 
  • #12
Nylex said:
[tex]\tan(x-30^{\circ}) = \tan 50^{\circ}[/tex]

Can you go from there?
Yup. Thanks!
 
  • #13
Zurtex said:
:smile: It's not that complex:
Hence my smilies
:wink:
 
  • #14
Kahsi said:
Hence my smilies
:wink:

Which if you expand is:

[tex]\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}= \frac{8 \cos^7 (10) \sin (10) - 56 \cos^5 (10) \sin^3 (10) + 56 \cos^3 (10) \sin^5 (10) - 8 \cos (10) \sin^7 (10)}{\cos^8(10) - 28 \cos^6(10) \sin^2(10) + 70 \cos^4(10) \sin^4(10) - 28 \cos^2 (10) \sin^6(10) + \sin^8 (10)}[/tex]

And it just so happens that nicely simplifies down to:

[tex]\frac{8 \cos^7 (10) \sin (10) - 56 \cos^5 (10) \sin^3 (10) + 56 \cos^3 (10) \sin^5 (10) - 8 \cos (10) \sin^7 (10)}{\cos^8(10) - 28 \cos^6(10) \sin^2(10) + 70 \cos^4(10) \sin^4(10) - 28 \cos^2 (10) \sin^6(10) + \sin^8 (10)} = \tan (80)[/tex]

:rolleyes:
 

Related to Finding the Angles: A Trigonometric Problem

1. What is the "Find all the angles problem"?

The "Find all the angles problem" is a mathematical problem that involves determining the measurements of all the angles in a given shape or figure. This can include various types of angles, such as acute, right, obtuse, and reflex angles.

2. How do I approach solving the "Find all the angles problem"?

To solve the "Find all the angles problem", you will need to use the properties and rules of angles, such as the sum of angles in a triangle or quadrilateral, and the relationships between angles formed by intersecting lines. You may also need to use tools like a protractor or ruler to accurately measure angles.

3. What are some common strategies for solving the "Find all the angles problem"?

Some common strategies for solving the "Find all the angles problem" include breaking the shape into smaller, simpler shapes, using known angles and their relationships to find unknown angles, and using algebraic equations to represent the relationships between angles.

4. What are some tips for avoiding mistakes when solving the "Find all the angles problem"?

To avoid mistakes when solving the "Find all the angles problem", make sure to carefully label and organize your work, double-check your calculations, and use multiple strategies to confirm your answers. It can also be helpful to work backwards or try different methods to check your work.

5. How is the "Find all the angles problem" relevant in real life?

The "Find all the angles problem" is relevant in various fields, such as architecture, engineering, and surveying, where accurate measurements of angles are crucial for designing and constructing buildings, roads, and other structures. It is also used in navigation and mapmaking to determine the direction and distance between two points.

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