The question that we are given a polygon on left on diagram with each side 20m. Then in the right side I made a triangle of ABC. I drew head and tail of the vector on my own. Angle I got between vector AB and BC is 60. Therefore I drew components of BC . Then the triangle (green colour) I considered as my triangle whose length AC I have to find.
My teacher got the answer 20$\sqrt{3}$. So what I did is that $20^2 + 10^2 + 10$$\sqrt{3}$ As sum of Base and perpendicular = $ H^2$
My sir answer is 20$\sqrt{3}$ but I am not getting the same answer.Where am I wrong ?
I have tagged vectors since I want to confirm whether the way I drew head and tail of vector and components is right or not.
When working with trigonometry for vectors, only the magnitudes come into play. As such, the direction of the vectors drawn do not make a difference in calculating magnitudes.
Now for your question, without drawing the outside right triangles, this problem can be solved by using the law of cosines $$c^2=a^2+b^2-2ab\cos\gamma,$$ where in your case, $a=b=AB=BC=20$, $c=AC$, and $\gamma=120^\circ$. Just showing the math here: \begin{align} AC&=\sqrt{20^2+20^2-2(20)(20)\cos120^\circ}\\ &=\sqrt{800+400}\\ &=\sqrt{1200}\\ &=20\sqrt3. \end{align}