# Finding the new force of vectors

It's been a while since the last time used vectors. I came across with the following question.

Find the Net force (size and orientation) of the vectors $$\vec{A}=15,37^{\circ},\,\vec{B}=5,162^{\circ} \; \text{and} \;\vec{C}=10,320^{\circ}.$$ The answers are $$\theta=15.557^{\circ} \; \text{and} \; R=15.45$$.

I have just copied the questions as is from the text-book above. I don't understand what $$\vec{A}=15,37^{\circ}$$ means. Is $$15$$ the size of the vector $$A$$? What does $$37^{\circ}$$ mean? Is it the angle between the vector $$A$$ and the next vector? I don't understand the input data.

• A note for voters: I don't think this question should be closed under our homework policy. In fact I think it's a pretty good example of how homework-like questions should be asked here. It presents a homework problem but is not asking for a solution to the problem; instead it's asking a specific conceptual question, in this case about the interpretation of part of the problem text. – David Z Mar 16 at 23:44
• What is the textbook? Isn't the notation explained in the textbook? – sammy gerbil Mar 17 at 0:02
• @sammygerbil It's a textbook with just exercises. It was also translated from other language, so I'm sorry if I'm not using the right expressions. – vesii Mar 17 at 9:57

The first number is the magnitude (i.e., length) of the vector. The second number is the direction of the vector in a plane, expressed as an angle measured from some fixed direction. The usual convention is to measure angles counterclockwise from the $$+x$$ direction.
I believe the notation $$\vec A=15,\ 37^\circ$$ is the polar coordinates of vector $$\vec A$$, where $$15$$ is the magnitude of vector $$\vec A$$, and $$37^\circ$$ is azimuth — the angle between the vector and some fixed ray. Since you are asked to find the azimuth of the result in the same coordinate system, it doesn't matter what this fixed ray is.
• Compute the $x$ and $y$ components of each of the three vectors. Add them to get the total vector. Compute its magnitude and direction from its components. – G. Smith Mar 17 at 19:49