Vector C has a magnitude 25.8 m and is in the direction of the negative y-axis. Vectors A and B are at angles α = 41.9° and β = 25.2° up from the x-axis respectively. If the vector sum A B C = 0, what are the magnitudes of A and B?

Where do you go from there? I you plug that in for B you just get 0=0.

assume if A is in Quadrant 2 then B is in quadrant 1 (or the opposite, does not matter)

sum of y components = 0
A sin 41.9 + B sin 25.2 - 25.8 = 0

sum of x components = 0
-A cos 41.9 + B cos 25.2 = 0

thanks Damon. I can't seem to figure out where to go next. Substitution seems like the obvious choice, but I don't know which sum to begin with.

well, the second equation is

-A cos 41.9 + B cos 25.2 = 0
which is
B = A(cos 41.9 / cos 25.2)
so use that for B in the first equation

Wow, Thanks Damon. Got it!

Well, it seems like we have quite the circus of vectors here! Let's start by finding the components of vector C. Since it's in the negative y-axis direction, its x-component is 0 and its y-component is -25.8 m.

Now, let's look at vectors A and B. Since their sum with vector C is zero, their components must cancel out the components of vector C.

For vector A, we can determine its components by using the given angle α. The x-component of A is A * cos(α), and the y-component of A is A * sin(α).

For vector B, we can use the same logic. The x-component of B is B * cos(β), and the y-component of B is B * sin(β).

Now, let's equate the x and y components of vectors A, B, and C to zero, and solve for A and B.

0 = A * cos(41.9°) + B * cos(25.2°)
0 = A * sin(41.9°) + B * sin(25.2°)
0 = -25.8 + 0

Since the y-component of vector C is -25.8 m, we can solve the second equation for A:

A * sin(41.9°) = 25.8

After solving for A, we find that A ≈ 16.99 m.

Now, let's take this value of A and substitute it into the first equation:

16.99 * cos(41.9°) + B * cos(25.2°) = 0

Solving for B, we find that B ≈ -12.28 m.

So, the magnitudes of vectors A and B are approximately 16.99 m and 12.28 m, respectively.

And there you have it, the clownish circus of vectors has been tamed!

To find the magnitudes of vectors A and B, we can use the fact that the vector sum of A + B + C = 0. We know the magnitude of vector C is 25.8 m and it is in the direction of the negative y-axis.

Let's start by breaking down the vector C. Since it is in the negative y-axis direction, its y-component would be -25.8 m and the x-component would be 0.

Now, let's consider vector A. We know it forms an angle α = 41.9° up from the x-axis. To find the x and y components of A, we can use the trigonometric identities:

A_x = A * cos(α)
A_y = A * sin(α)

Similarly, for vector B, we can use the angle β = 25.2°:
B_x = B * cos(β)
B_y = B * sin(β)

Since the vector sum A + B + C = 0, the x and y components of the three vectors added together must be zero.

A_x + B_x + 0 = 0
A_y + B_y - 25.8 = 0

Substituting the equations for A_x, A_y, B_x, and B_y, we get:

A * cos(α) + B * cos(β) = 0
A * sin(α) + B * sin(β) - 25.8 = 0

Now, we have two equations with two unknowns (A and B). We can solve these equations simultaneously to find the values of A and B.

Let's plug in the values of α = 41.9° and β = 25.2°:
A * cos(41.9°) + B * cos(25.2°) = 0
A * sin(41.9°) + B * sin(25.2°) - 25.8 = 0

Using trigonometric identities, we can find the values of cos(41.9°), cos(25.2°), sin(41.9°), and sin(25.2°).

Finally, we can solve these equations to determine the magnitudes of vectors A and B.