Vector addition problem - component method

[Larrytsai]

Vplane = 63km/hr at x degrees and direction
Vwind = 47km/hr at 211 degrees
Vg = unknown speed at 165 degrees

find the unknowns


my attempt using component method:

(Cos 165)Vg = (Cos 211)47 + (Cos theta)63

(Sin 165)Vg = (Sin 211)47 + (Sin theta)63

from here i think I am supposed to isolate Vg and then subsitute to find the unknown angle and then plug the angle into get the speed but I am not quite sure

I know i can use the cosine law or sine law but I have to learn both ways so that's why I am using the component method

 

[gabbagabbahey]

Vplane = 63km/hr at x degrees and direction
Vwind = 47km/hr at 211 degrees
Vg = unknown speed at 165 degrees

find the unknowns


my attempt using component method:

(Cos 165)Vg = (Cos 211)47 + (Cos theta)63

(Sin 165)Vg = (Sin 211)47 + (Sin theta)63

from here i think I am supposed to isolate Vg and then subsitute to find the unknown angle and then plug the angle into get the speed but I am not quite sure

I know i can use the cosine law or sine law but I have to learn both ways so that's why I am using the component method

I think the easiest way for you to solve this system of equations is to isolate cos(\theta) in the first equation, sin(\theta) in the second one, and then square both equations, add them together and take advantage of the fact that sin^2(\theta)+cos^2(\theta)=1 to sollve for V_g. After that, plug your solution into either equation and solve for \theta. If you post your solution, I'll be happy to check it.

 

[Larrytsai]

hmmm, that's where i ran into trouble, isolating theta i don't understand how to isolate theta. like do i use Sin-1? and how would it look?

 

[gabbagabbahey]

hmmm, that's where i ran into trouble, isolating theta i don't understand how to isolate theta. like do i use Sin-1? and how would it look?

Start with the first equation:
V_gcos(165^{\circ}) = 47cos(211^{\circ}) + 63cos(\theta)

Can you solve this for cos(\theta)?

 

[Larrytsai]

ohhh okay so now i got (Vg(Cos 165) - 47(Cos 211))/63 = Cos (theta)

i think that's right

 

[gabbagabbahey]

ohhh okay so now i got (Vg(Cos 165) - 47(Cos 211))/63 = Cos (theta)

i think that's right

Yup, so what is cos^2(\theta) then?

 

[Larrytsai]

by Cos^2 u mean Cos-1 right? and if so i don't see what will happen to my other unknown

 

[gabbagabbahey]

by Cos^2 u mean Cos-1 right? and if so i don't see what will happen to my other unknown

Nope, I mean (cos(\theta))^2...it will become clear why I want you to find this in a minute.

 

[Larrytsai]

oo I am just lost now sorrry =(

 

[gabbagabbahey]

oo I am just lost now sorrry =(

If
cos(\theta)=\frac{V_gcos(165^{\circ})-47cos(211^{\circ})}{63}
then
cos^2(\theta)=\frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}

...Now, solve the second equation for sin(\theta) and then find sin^2(\theta)

 

[Larrytsai]

okay so i got

Sin^2(theta) = ((Vg(Sin165)-47(Sin211))^2)/(63)^2

well now I am wondering does that Cos^2 and Sin^2 isolate theta?

 

[gabbagabbahey]

okay so i got

Sin^2(theta) = ((Vg(Sin165)-47(Sin211))^2)/(63)^2

well now I am wondering does that Cos^2 and Sin^2 isolate theta?

Okay, so now you have:
cos^2(\theta)=\frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}
and
sin^2(\theta)=\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2}

so what is sin^2(\theta)+cos^2(\theta)?

 

[Larrytsai]

from what u said earlier it = 1

 

[gabbagabbahey]

from what u said earlier it = 1

Yes, there is a trigonometric identity: http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity" that it will equal 1 for any theta.

so you get:
sin^2(\theta)+cos^2(\theta)=\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2} + \frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}=1
And so,
\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2} + \frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}=1

Now, can you expand this equation to find a quadratic equation for V_g?

 

[Larrytsai]

okay so i tried to find wat Vg is equal to and i get 2.11 which seems wrong to me

my work is shown here

1=((Vg(cos165)+40.287)^2 + (Vg(Sin165)+24.207)^2)/(3969)
multiply both sides by 3969

3969=(Vg(Cos165)+40.287)^2 + (Vg(Sin165)+24.207)^2
square root both sides

63=-0.7071Vg+64.494
subtract 64 to both sides

-1.494= -0.7071Vg

0.7072 came from cos165+sin165

Vg = -1.494/-0.7071
Vg= 2.11

 

[gabbagabbahey]

3969=(Vg(Cos165)+40.287)^2 + (Vg(Sin165)+24.207)^2
square root both sides

you can't just take the square root of the right hand side here, because it is a sum of 2 squares not just a perfect square...you have to expand each term first like this:


\frac{[V_gsin(165^{\circ})-47sin(211^{\circ})]^2}{63^2} + \frac{[V_gcos(165^{\circ})-47cos(211^{\circ})]^2}{63^2}=1


\Rightarrow [V_gsin(165^{\circ})-47sin(211^{\circ})]^2 +[V_gcos(165^{\circ})-47cos(211^{\circ})]^2=63^2



\Rightarrow [V_g^2sin^2(165^{\circ})-94V_gsin(165^{\circ})sin(211^{\circ})+47^2sin^2(211^{\circ})] +[V_g^2cos^2(165^{\circ})-94V_gcos(165^{\circ})cos(211^{\circ})+47^2cos^2(211^{\circ})]=63^2



\Rightarrow [sin^2(165^{\circ})+cos^2(165^{\circ})]V_g^2-94V_g[sin(165^{\circ})sin(211^{\circ})+cos(165^{\circ})cos(211^{\circ})]+47^2[sin^2(211^{\circ})] +cos^2(211^{\circ})]=63^2

Now, there are two useful Trig Identities: (1) sin^2(x)+cos^2(x)=1 and (2) sin(x)sin(y)+cos(x)cos(y)=cos(y-x).
Using these you get:

(1)V_g^2-94cos(211^{\circ}-165^{\circ})V_g+(1)47^2=63^2


\Rightarrow V_g^2-94cos(46^{\circ})V_g+(47^2-63^2)=0

Now, can you solve for V_g?

You should get 2 solutions!