Imagine that we have a vector v=ai+bj. In the coordinate grid it is represented as an arrow that starts at (0,0) and ends at (a,b):
As you can see, this vector along with its x and y components (a and b) form a right triangle. Applying the Pythagorean theorem to this triangle results in:
[tex]\parallel v\parallel=\sqrt[]{a^2+b^2}[/tex]
Now, remember that the tangent of an angle in a right triangle is given by the quotient between its opposite side and its adjacent side. As you can see the opposite and adjacent side of angle α are b and a respectively which means that its tangent is equal to:
[tex]\tan \alpha=\frac{b}{a}[/tex]
We can apply all of this to the vector given by the question. We have the value of ||v|| and we have the value of the angle (in this case is named θ instead of α). With these two values we can build two equations for a and b:
[tex]\begin{gathered} 36=\sqrt[]{a^2+b^2} \\ \tan 120^{\circ}=\frac{b}{a} \end{gathered}[/tex]
Let's take the second equation and multiply both sides by a:
[tex]\begin{gathered} a\cdot\tan 120^{\circ}=\frac{b}{a}\cdot a \\ b=a\cdot\tan 120^{\circ}=-a\sqrt[]{3} \end{gathered}[/tex]
So we have an expression for b. We use it in the first equation:
[tex]\begin{gathered} 36=\sqrt[]{a^2+b^2}=\sqrt[]{a^2+(-a\sqrt[]{3})^2} \\ 36=\sqrt[]{a^2+(-a\sqrt[]{3})^2}=\sqrt[]{a^2+3a^2}=\sqrt[]{4a^2}=2|a| \\ 36=2|a| \end{gathered}[/tex]
We divide both sides by 2:
[tex]\begin{gathered} \frac{2|a|}{2}=\frac{36}{2} \\ |a|=18 \end{gathered}[/tex]
Then we apply module to the expression for b and use the value |a|=18:
[tex]|b|=|-a\sqrt[]{3}|=|a|\sqrt[]{3}=18\cdot\sqrt[]{3}[/tex]
So we have the modules of a and b, we still need to decide their sign:
[tex]\begin{gathered} |a|=18 \\ |b|=18\sqrt[]{3} \end{gathered}[/tex]
In order to do this we can use the value of the angle θ. It's 120° which means that it's in the second quadrant because is greater than 90° and smaller than 180°. If the vector is this quadrant then its i component is negative and its j component is positive which means a<0 and b>0. Then we get a=-18 and b=18√3 so we get:
[tex]v=-18i+18\sqrt[]{3}j[/tex]
So the answers are the following two numbers:
[tex]-18,18\sqrt[]{3}[/tex]
IMPORTANT: In case you can't write radicals you must replace the square root of 3 with 1.732 so if you can't write radicals the answes are:
[tex]-18,31.177[/tex]
