Solving -25FCosα + 1.5FSinα= -80 with Trigonometric Identities

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Homework Help Overview

The discussion revolves around the equation -25FCosα + 1.5FSinα = -80, which involves trigonometric identities and the challenge of solving for the angle α given the force F. The context is within trigonometric equations and their manipulation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the solvability of the equation with two unknowns and discuss potential methods for solving trigonometric equations. Some suggest using substitutions for cosine and sine, while others propose iterative methods to find the angle α based on a given value for F.

Discussion Status

The discussion is active, with participants offering various approaches to tackle the problem. There is recognition of the complexity due to the number of unknowns and the need for clarification on the units involved. Iterative methods are suggested as a possible way to approximate the solution for α.

Contextual Notes

Participants note the importance of the value of F, which is specified as 4 kN, but there remains uncertainty regarding the units of the constant -80. This ambiguity may affect the interpretation of the equation.

Esas Shakeel
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Poster has been reminded to post schoolwork in the Homework Help forums & show and attempt at a solution
-25FCosα + 1.5FSinα= -80

Can someone please solve this and tell what trigonometric identities are we going to be using this solving?
 
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There are two unknowns in one equation, doesn't seem to be solvable.
 
Esas Shakeel said:
-25FCosα + 1.5FSinα= -80

Can someone please solve this and tell what trigonometric identities are we going to be using this solving?
I can't think of any. For a given value of F, your best bet would be to iterate to find the angle α which satisfies this equation.
 
You can try with some methods for solving trigonometric equations... For example putting ##X=\cos{\alpha}## and ##Y=\sin{\alpha}## you can form the system ##\left\{ \begin{array}{rl} -25FX+1.5FY=-80 \\ X^2+Y^2=1 \end{array} \right.##
 
In general, an equation ##a\sin(\alpha) + b\cos(\alpha)=c## can be rewritten as
##sin(\alpha + \beta)=\frac{c}{\sqrt{a²+b²}}##,
where ##\beta## satisfies ##\cos(\beta)=\frac{a}{\sqrt{a²+b²}}##, ##\sin(\beta)=\frac{b}{\sqrt{a²+b²}}##.
 
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Ssnow said:
You can try with some methods for solving trigonometric equations... For example putting ##X=\cos{\alpha}## and ##Y=\sin{\alpha}## you can form the system ##\left\{ \begin{array}{rl} -25FX+1.5FY=-80 \\ X^2+Y^2=1 \end{array} \right.##
That still gives three unknowns with two equations.
 
blue_leaf77 said:
That still gives three unknowns with two equations.
But F = 4 kN, according to the OP. It's still not clear if 80 is in kN or what.

Knowing a value for F, you can still solve the original equation by iterating for the angle α.

-100 kN ⋅ cos α + 6 kN ⋅ sin α = -80 kN (?)

f(α) = 80 - 100 ⋅ cos α + 6 ⋅ sin α

Code:
   α         f(α)
  Deg.       kN
  10       -17.44
  15       -15.04
  20       -11.92
  25        -8.10
  30        -3.60
  35        +1.53
α lies somewhere between 30° and 35°.

You can continue the iteration to reach the desired precision for α.
 
Yes @blue_leaf77, it will be a system with three unknowns and two equations, or a system with two unknowns, one parameter ##F## and two equations :-D
 

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