Solve 400sinA + 500cosA = 600: Tips & Tricks

[strokebow]

Can anyone tell me how to solve...

400sinA + 500cosA = 600

Just doing some statics problems and for one of 'em I needed to find A. Can anyone tell me how its done.

Thanks

 

[Eighty]

See this recent thread: https://www.physicsforums.com/showthread.php?t=164679

 

[HallsofIvy]

First you need to know that sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B) for any A, B. Because we obviously can't have cos(B)= 400 and sin(B)= 500 (sin and cos values are always <= 1), multiply that entire equation by C: Csin(A+ B)= Csin(A)cos(B)+ Ccos(A)sin(B). You want to find C and B so that C cos(B)= 400 and C sin(B)= 500. Since sin²(B)+ cos²(B)= 1, C² cos²(B)+ C²sin²(B)= C²= (400)²+ (500)²= 160000+ 250000= 410000. C²= 410000 so $C= \sqrt{410000}= 100\sqrt{41}$. Now you know that [itex]C cos(B)= 100\sqrt{41}cos(B)= 400 so $cos(B)= 4/\sqrt{41}$. $B= arccos(4/\sqrt{41})$ so we now have
$$100\sqrt{41} sin(A+ arccos(4/\sqrt{41})= 600$$
$$sin(A+ arccos(4/\sqt{41})= 6/\sqrt{41}$$
so $A+ arccos(4/\sqrt{41})= arcsin(6/\sqrt{41})$ and, finally,
$$A= arcsin(6/\sqrt{41})- arccos(4/\sqrt{41})$$
According to my calculator, that is about 0.318.

 

[robert Ihnot]

Using $$sin(x)=\sqrt{1-cox^2(x)}$$ I arrived also at a second answer, x=1.0315.

 

[waht]

I'm not sure your static problem would lead to that equation. Sum of forces in x direction equal 0 and likewise in y. Your equation implies adding an x component and a y component. Once again I don't know what the problem is so I could be wrong.

 

[Werg22]

I'm not sure your static problem would lead to that equation. Sum of forces in x direction equal 0 and likewise in y. Your equation implies adding an x component and a y component. Once again I don't know what the problem is so I could be wrong.

Huh, I think he is adding moments of forces.