Two forces act at a point. F_A = 700 N points 30 degrees above the positive x-axis. F_B has unknown magnitude and makes an angle theta with the negative x-direction, so its horizontal component points the opposite way from F_A. Find the magnitude of F_B and the angle theta so that the resultant is 1500 N directed along the positive y-axis.

Step-by-step solution

1. 1

   Turn the picture into two scalar conditions. A vector equation in the plane is two equations. "The resultant points along $+y$" says the total $x$-component is zero; "the resultant is $1500$ N" then says the total $y$-component is $1500$:

   $\sum F_x=0,\qquad \sum F_y=1500$

   This is the whole trick — you never need the resultant's direction cosines, only these two lines.

2. 2

   Resolve the known force. With $F_A$ at $30^\circ$ to the positive $x$-axis:

   $F_{Ax}=700\cos 30^\circ=606.2178\text{ N},\qquad F_{Ay}=700\sin 30^\circ=350\text{ N}$

   Keep the extra decimals in $F_{Ax}$; rounding it to $606$ here moves the final angle by about a tenth of a degree.

3. 3

   Apply the two conditions to $F_B$. Writing $F_B$'s components as $-F_B\cos\theta$ (opposing $F_A$) and $+F_B\sin\theta$:

   $606.2178-F_B\cos\theta=0\;\Longrightarrow\;F_B\cos\theta=606.2178$
   $350+F_B\sin\theta=1500\;\Longrightarrow\;F_B\sin\theta=1150$

   Two equations, two unknowns — but they are not linear, so solve them as a pair.

4. 4

   Square and add to get the magnitude. Squaring both equations and adding kills $\theta$ because $\sin^2\theta+\cos^2\theta=1$:

   $F_B^2=1150^2+606.2178^2=1322500+367500=1690000$

   $F_B=\sqrt{1690000}=1300\text{ N}$

   The numbers are exact here: $606.2178^2=367500$ to the digit, so $F_B$ comes out as a clean $1.30\times 10^3$ N.

5. 5

   Divide to get the angle. Dividing the second equation by the first removes $F_B$:

   $\tan\theta=\frac{1150}{606.2178}=1.89701\;\Longrightarrow\;\theta=\arctan(1.89701)=62.2^\circ$

   (Carrying $606.2$ instead of the exact $700\cos 30^\circ$ is what produces the commonly seen $62.3^\circ$; the correct value to one decimal is $62.2^\circ$.)

6. 6

   Check the answer against both conditions. With $F_B=1300$ and $\theta=62.204^\circ$:

   $1300\cos 62.204^\circ=606.2\quad\text{cancels }F_{Ax}\;\checkmark$
   $350+1300\sin 62.204^\circ=350+1150=1500\;\checkmark$

   Both conditions hold, so the resultant really is $1500$ N straight up the $y$-axis.