# How do I solve this?

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1. Feb 19, 2016

### Esas Shakeel

• 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?

2. Feb 19, 2016

### blue_leaf77

There are two unknowns in one equation, doesn't seem to be solvable.

3. Feb 19, 2016

### SteamKing

Staff Emeritus
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.

4. Feb 19, 2016

### Ssnow

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 $\left\{ \begin{array}{rl} -25FX+1.5FY=-80 \\ X^2+Y^2=1 \end{array} \right.$

5. Feb 19, 2016

### Esas Shakeel

Im so sorry i forgot to mention the value for F! Its 4kN
@Ssnow @SteamKing

6. Feb 19, 2016

### Samy_A

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 $\cos(\beta)=\frac{a}{\sqrt{a²+b²}}$, $\sin(\beta)=\frac{b}{\sqrt{a²+b²}}$.

7. Feb 19, 2016

### blue_leaf77

That still gives three unknowns with two equations.

8. Feb 19, 2016

### SteamKing

Staff Emeritus
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 (Text):

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.

9. Feb 19, 2016

### Ssnow

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