Solving for X: pCosX & pSinX Equations

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To solve the equations pCosX = 6 and 0.2(pSinX + 25) = 6, first isolate pSinX, resulting in pSinX = 5. The discussion clarifies that "p" is a constant, not a variable. A suggested method for finding X involves rearranging the equations, squaring them, and then adding the results. This approach leverages the relationship between sine and cosine to simplify the problem. The conversation emphasizes understanding the constants involved in the equations for accurate solutions.
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i know that pCosX = 6 and 0.2(pSinX+25)= 6

pSinx = 5
pCosx = 6

how do i solve for X.. I'm not sure how this works
 
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What do you mean by psin(x)? What does the p stands for? Is it a variable?
 
nvm... its constant
 
DeanBH said:
i know that pCosX = 6 and 0.2(pSinX+25)= 6

pSinx = 5
pCosx = 6

how do i solve for X.. I'm not sure how this works

Hi DeanBH! :smile:

Hint: whenever you have cos and sin (and nothing else), rearrange, take squares, and add. :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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