# Vector addition problem - component method

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 im supposed to isolate Vg and then subsitute to find the unknown angle and then plug the angle in to get the speed but im not quite sure

I know i can use the cosine law or sine law but I have to learn both ways so thats why im using the component method

## Answers and Replies

gabbagabbahey
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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 im supposed to isolate Vg and then subsitute to find the unknown angle and then plug the angle in to get the speed but im not quite sure

I know i can use the cosine law or sine law but I have to learn both ways so thats why im 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.

hmmm, thats where i ran in to trouble, isolating theta i dont understand how to isolate theta. like do i use Sin-1? and how would it look?

gabbagabbahey
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hmmm, thats where i ran in to trouble, isolating theta i dont understand how to isolate theta. like do i use Sin-1? and how would it look?

$$V_gcos(165^{\circ}) = 47cos(211^{\circ}) + 63cos(\theta)$$

Can you solve this for $$cos(\theta)$$?

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

i think thats right

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

i think thats right

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

by Cos^2 u mean Cos-1 right? and if so i dont see what will happen to my other unknown

gabbagabbahey
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by Cos^2 u mean Cos-1 right? and if so i dont 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.

oo im just lost now sorrry =(

gabbagabbahey
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oo im 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 $$sin^2(\theta)$$

okay so i got

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

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

gabbagabbahey
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okay so i got

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

well now im 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 $$sin^2(\theta)+cos^2(\theta)$$?

from what u said earlier it = 1

gabbagabbahey
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from what u said earlier it = 1

Yes, there is a trigonometric identity: http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity" [Broken] 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 $$V_g$$?

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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
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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!

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