This is a homework problem which has me stumped. I have just begun Calculus 3 and this is the first section introducing vectors. Problem 46 from section 12.2 of Thomas' Calculus 12th edition.

The embedded picture states the problem and shows some of my hen's scratching in an effort to solve it. After studying the example problem from the same sub-chapter, I understand that the two vectors given, when summed, is the vector that is acting in opposition to the vector pulling the weight down. That is $\mathbf{F_1} + \mathbf{F_2} = \langle 0,50 \rangle$. This leads to the following system of equations:

$$ \begin{array}{rcl} -\left|\mathbf{F_1}\right|cos(\alpha) + \left|\mathbf{F_2}\right|cos(60^\circ) & = & 0 \\ \left|\mathbf{F_1}\right|sin(\alpha) + \left|\mathbf{F_2}\right|sin(60^\circ) & = & 50 \\ \end{array} $$

Basically, I'm having troubles figuring out how to eliminate one of the two unknowns: $\alpha$ and $\left|\mathbf{F_2}\right|$. Using the system of equations, I can tell you that $\left|\mathbf{F_2}\right|$ is as follows:

$$ \begin{array}{rcl} -\left|\mathbf{F_1}\right|cos(\alpha) + \left|\mathbf{F_2}\right|cos(60^\circ) & = & 0 \\ -35cos(\alpha) + \left|\mathbf{F_2}\right|\frac{1}{2} & = & 0 \\ -70cos(\alpha) + \left|\mathbf{F_2}\right| & = & 0 \\ \left|\mathbf{F_2}\right| & = & 70cos(\alpha) \\ \end{array} $$

Also, knowing that the interior angles of a triangle must sum $180^\circ$, it's also quite a simple matter to demonstrate that $\alpha$ is as follows (assigning $\beta$ to the unnamed angle):

$$ \begin{array}{rcl} 180 & = & 60 + \alpha + \beta \\ \alpha & = & 120 - \beta \\ \end{array} $$

From the scratching notes you can see in the picture, you can see that I've tried to draw in the unseen vector $\langle 0,50 \rangle$ in the hopes of making a right triangle to get myself closer to the answer. This didn't help me. I had the inspiration that some of the trigonometric functions and identities I learned (too many years ago now) would be of use. So I pulled out my pre-calc book to look up things like SSA and so forth. However, none seem to be the answer. For example, the SSA only works if you know 2 of the three sides and one angle. Well, I've got to of the three required criteria.

I also refreshed myself on the law of sines. However, this doesn't bring me closer to (what I think are) the solutions. So, one again, I can tell you this much:

$$ \text{The law of sines} \\ \frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c} \\ \text{I don't know C, or c, but this shouldn't matter} \\ \Rightarrow \frac{sin(60)}{35} = \frac{sin(\alpha)}{\left|\mathbf{F_2}\right|} $$

However, it's pretty obvious that, though I can calculate the left side, it doesn't bring me close enough to get either $\alpha$ or $\left| \mathbf{F_2} \right|$

So, if someone could kindly point me in the right direction, I'd be very grateful. Thanks.

  • 1
    $\begingroup$ Why don't you use $|F_2|=70\cos(\alpha)$ in $|F_1|sin(α)+|F_2|\frac{\sqrt3}{2} = 50$ then you have one equation with one var :) $\endgroup$
    – Lu_kors
    Jan 19, 2015 at 0:58
  • $\begingroup$ @Lu_kors I had thought of that, but quickly rejected it because I still had two variables: right? I don't know what $\alpha$ is and thus don't know $sin(\alpha)$ or $cos(\alpha)$ in addition to not knowing $\left| \mathbf{F_2} \right|$. What am I forgetting that you obviously see? $\endgroup$ Jan 19, 2015 at 18:06

2 Answers 2


Using Lami's there

$\frac{F_2}{sin (90+\alpha)}=2F_1=\frac{50}{sin (120-\alpha)}$

$\frac{F_2}{cos \alpha}=2F_1=\frac{50}{sin (120-\alpha)}$

$\frac{F_2}{cos \alpha}=\frac{50}{sin 120cos\alpha-cos 120sin\alpha}$

$\frac{F_2}{cos \alpha}=\frac{50}{\frac{\sqrt3}{2}cos\alpha+\frac{1}{2}sin\alpha}$

$\frac{F_2}{cos \alpha}=\frac{100}{\sqrt3cos\alpha+sin\alpha}$

Substituting your answer above where you got $F_2=70 cos \alpha$,



By using double angle,

R=2, $\theta=60$






Since $60\le \alpha \le 120, \alpha=-14.4$


I am getting the angle to be -ve, meaning sin is -ve and cos is +ve in the 4th quadrant. Angles in the 3rd quadrant too keep the sin -ve exceeds 120. It sounds a bit improbable to have a negative angle in real life, but according to the question, it makes sense.

Good luck.


Use $|F_2|=70\cos(\alpha)$ in $|F_1|sin(α)+|F_2|\frac{\sqrt3}{2} = 50$ then you have:

$$ 35(\sin{\alpha} + \sqrt{3} \cos\alpha)=0$$ Divide by $35\cos\alpha$ $$ \tan\alpha = -\sqrt3$$ So you can find $\alpha = -60°$

No responsibility for possible calc mistakes :P

  • $\begingroup$ I'm sorry but I don't see where you get $35(sin\alpha + \sqrt{3} \cdot cos\alpha) = 0$ $\endgroup$ Jan 19, 2015 at 22:40
  • $\begingroup$ Ok I see where the left-hand side comes from, but where do you get 0 on the right? $\endgroup$ Jan 19, 2015 at 22:47
  • $\begingroup$ jup sry has to be 50, will edit it, asap i have a bit more time $\endgroup$
    – Lu_kors
    Jan 19, 2015 at 23:14
  • $\begingroup$ I should have marked this as answered back then. Sorry, and thanks for the help. $\endgroup$ Jan 13, 2020 at 19:15

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