Two unknowns in trigonometric calculation?

In summary, the conversation discusses a problem involving two boats leaving from different harbors at different speeds and trying to determine the course the second boat should take to meet the first. The person discussing the problem considers various equations, such as the definitions of sine, cosine, and tangent, the law of sines and cosines, the Pythagorean theorem, and the area formula. They also describe their attempts at solving the problem using these equations, but ultimately find a solution by finding the value of x and using it to find v.
  • #1
Ricaoma
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Ricaoma said:
Hah, nevermind, I found a way to find x and then I could just insert the value to get v, just like I thought! :biggrin:

Homework Statement


A boat leaves the harbor going 60 degrees north-east with the speed 20km/h. 3 hours later, another boat leaves a harbor 160km to the east with a speed of 70km/h. What course should the second boat have to meet the first?

Homework Equations


Not 100% sure what equations would be relevant, but those who've crossed my mind have been:

The definitions of sine, cosine, tangent
[tex]\sin A = \frac{opposite}{hypotenuse}[/tex]
[tex]\cos A = \frac{adjacent}{hypotenuse}[/tex]
[tex]\tan A = \frac{opposite}{adjacent}[/tex]

Law of sines
[tex]\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}[/tex]

Law of cosines
[tex]c^2 = a^2 + b^2 - 2ab\cos C[/tex]

Pythagorean theorem
[tex]a^2 + b^2 = c^2[/tex]

Area
[tex]A = \frac{a*b*\sin C}{2}[/tex]

The Attempt at a Solution


My sketched figure is attached to this post.

Start-phase
I started out by sketching a triangle with the base 160, the left side being 20t (20*time in hours since the first boat left) and the right being 70t-210 (70*time in hours since the first boat left - 70*3 for the 3 hours between the starting time).

Splitting the triangle
I also drew a line in the middle to make two right-angled triangles. Knowing that the bottom-left angle was 60 degees and that the new angle was 90 degrees I subtracted those two values from 180 to get the top angle of the left right-angled triangle.
[tex]180-(60+90) = 30[/tex]

After that I used the definition of sine to calculate the length of the middle line, like this:
[tex]\sin60 = \frac{h}{20t}[/tex]
[tex]20t * \sin60 = \frac{h}{20t} * 20t[/tex]
[tex]20t * \sin60 = h[/tex]

I did the same thing to find out the base of the left triangle:
[tex]\sin30 = \frac{b}{20t}[/tex]
[tex]20t * \sin30 = \frac{b}{20t} * 20t[/tex]
[tex]20t * \sin30 = b[/tex]

With this knowledge I know that the base of the right-side triangle's base is
[tex]160 = 20t*sin(30) + b2[/tex]
[tex]160 - 20t * \sin30[/tex]

The problem
I've tried all the things I've come up with so far, and most of them gives the same result. What I have to do is to get t out of any of the equations and then use that value to get rid of the t in all the equations. However, I have not yet been able to find an equation where this is an easy/possible task with my current trigonometric knowledge, which is pretty limited to the relevant equations listed above. Although, keep in mind that I may simply have missed an oppertunity to simplify an equation of overlooked some way of solving this, so the answer may still be in something that I think I've tried.

Some things I've tried:

Sine
[tex]\sin V = \frac{20t*\sin60}{70t-210}[/tex]

Cosine
[tex]\cos V = \frac{160-20t*\sin30}{70t-210}[/tex]

Law of Sine
Same result as my Sine-based approach
[tex]\frac{\sin90}{70t-210} = \frac{\sin V}{20t*\sin60}[/tex]
[tex]\sin V = \frac{20t*\sin 60}{70t-210}[/tex]

Law of Cossine
[tex](20t)^2 = (70t-210)^2 + 160^2 - 2*160*(70t-210)*\cos V[/tex]
[tex]\cos V = \frac{225t^2 - 1470t + 3485}{1120t - 3360}[/tex]

Equation system based on area
Same result as my Sine-based approach
[tex]A = \frac{a*b*\sin V}{2}[/tex]
[tex]A = \frac{a*h}{2}[/tex]
[tex]\frac{a*b*\sin V}{2} = \frac{a*h}{2}[/tex]
[tex]a*b*\sin V = a*h[/tex]
[tex]b*\sin V = h[/tex]
[tex]\sin V = \frac{h}{b}[/tex]
[tex]\sin V = \frac{20t*\sin60}{70t-210}[/tex]

Equation system based on Cosine and Law of Cosine
Don't know how to solve a third-degree equation like this
[tex]\cos V = \frac{160-20t*\sin30}{70t-210}[/tex]
[tex]\cos V = \frac{225t^2 - 1470t + 3485}{1120t - 3360}[/tex]
[tex]\frac{160-20t*\sin30}{70t-210} = \frac{225t^2 - 1470t + 3485}{1120t - 3360}[/tex]
[tex]315t^3 - 2769t^2 + 6797t - 3885 = 0[/tex]I'm really out of ideas now, so any ideas or thoughts are very much appreciated.
 

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  • #2
Hah, nevermind, I found a way to find x and then I could just insert the value to get v, just like I thought! :biggrin:
 

FAQ: Two unknowns in trigonometric calculation?

1. What are the two unknowns in trigonometric calculations?

The two unknowns in trigonometric calculations are typically the angle (represented by the Greek letter theta) and the side length of a right triangle.

2. How do I solve for two unknowns in a trigonometric equation?

To solve for two unknowns in a trigonometric equation, you need to have at least two equations with two unknowns each. You can then use algebraic methods such as substitution or elimination to solve for the unknowns.

3. Can I use trigonometry to solve for unknowns in non-right triangles?

Yes, you can use trigonometry to solve for unknowns in non-right triangles. However, you will need to use advanced trigonometric identities and formulas such as the law of sines and the law of cosines.

4. What are some common applications of solving for two unknowns in trigonometric calculations?

Solving for two unknowns in trigonometric calculations is commonly used in fields such as engineering, physics, and navigation. It can be used to determine the height of a building, the distance of a ship from shore, or the trajectory of a projectile.

5. Are there any special cases when solving for two unknowns in trigonometric calculations?

Yes, there are special cases such as when there is no solution or when there are infinitely many solutions. This can occur when the given information is contradictory or when the equations are equivalent. In these cases, it is important to check for extraneous solutions and to verify the solutions using the given information.

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