Really simple system of equations

In summary, the conversation discusses two equations (2500-Tac cosφ=0 and Tac sinφ-500=0) and the goal of solving for φ to find Tac. The user has gotten to the point of Tac=500/sinφ and 2500-500cosφ/sinφ=0, but is unsure of how to proceed to obtain the solution of 11.31°.
  • #1
yoleven
78
1

Homework Statement


1. 2500-Tac cos [tex]\Phi=0[/tex]
2. Tac sin[tex]\Phi-500=0[/tex]

Homework Equations





The Attempt at a Solution


I have 2 equations. I need to solve for [tex]\Phi[/tex] to get Tac. I have got to the point of :
Tac=500/sin[tex]\Phi[/tex]
2500-500/sin[tex]\Phi*(cos\Phi)=0[/tex]

after this i can't go any further

I am looking at the solution of 11.31°. But I need to solve to get it.
 
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  • #2
yoleven said:
I have got to the point of :
Tac=500/sin[tex]\Phi[/tex]
2500-500/sin[tex]\Phi*(cos\Phi)=0[/tex]

after this i can't go any further

I am looking at the solution of 11.31°. But I need to solve to get it.

Hi yoleven! :smile:

(have a phi: φ :smile:)

Your second line should be 2500 - 500cosφ/sinφ = 0. :wink:
 
  • #3


As a scientist, it is important to understand and use mathematical equations to solve problems. In this case, we have a system of equations with two unknowns, Tac and \Phi. To solve for these variables, we can use substitution or elimination methods.

Substitution method:
From the first equation, we can express Tac in terms of \Phi as Tac = 2500/cos\Phi. Substituting this value into the second equation, we get 2500/cos\Phi * sin\Phi - 500 = 0. Simplifying, we get 2500*tan\Phi - 500 = 0. Solving for tan\Phi, we get tan\Phi = 500/2500 = 0.2. Using a calculator, we can find the inverse tangent of 0.2, which is approximately 11.31°.

Elimination method:
Multiplying the first equation by sin\Phi and the second equation by cos\Phi, we get 2500sin\Phi - Tac = 0 and Tacsin\Phi - 500cos\Phi = 0. Adding these two equations, we get 2500sin\Phi - 500cos\Phi = 0. Dividing both sides by 500, we get sin\Phi - cos\Phi = 0. Using the trigonometric identity sin^2\Phi + cos^2\Phi = 1, we can rewrite this equation as sin\Phi = cos\Phi. Taking the inverse sine of both sides, we get \Phi = 45°. However, we need to check if this is a valid solution by substituting it into the original equations. We can see that it does not satisfy the second equation. Therefore, we need to use the first method to find the correct solution of approximately 11.31°.

In conclusion, we can solve this system of equations using different methods and obtain the same solution of approximately 11.31° for \Phi. This shows the importance of understanding and using mathematical equations in scientific research.
 

FAQ: Really simple system of equations

What is a really simple system of equations?

A really simple system of equations is a set of two or more equations that have two or more unknown variables, and each equation can be solved for one of the unknown variables.

How do you solve a really simple system of equations?

To solve a really simple system of equations, you can use the substitution method, elimination method, or graphing method. These methods involve manipulating the equations to eliminate one variable and solve for the other.

Why are really simple systems of equations important?

Really simple systems of equations are important because they can be used to model and solve real-world problems, such as finding the cost of items or determining the number of solutions to a problem.

Can a really simple system of equations have more than one solution?

Yes, a really simple system of equations can have one, infinite, or no solutions. This depends on the specific equations in the system and their relationship to each other.

What are some applications of really simple systems of equations in science?

Really simple systems of equations are commonly used in science to model and solve problems in areas such as physics, chemistry, biology, and engineering. They can be used to analyze relationships between variables and make predictions about different phenomena.

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