How do you find the angle between vector $\vec b-\vec a$ and the $x$-axis given that $\vec a$ has a length equal to $1$ and an angle of $10^\circ$ with the $x$-axis and $\vec b$ has a length of $1$ and an angle equal to $40^\circ$ with the $x$-axis?
The use of a calculator is forbidden so you can't just calculate the coordinates of $\vec a – \vec b$ simply by entering $(\cos 40^\circ – \cos 10^\circ, \sin 40^\circ – \sin 10 ^\circ)$ in a calculator. So what other approaches can I take?
Best Answer
HINT:
Draw a triangle $\triangle{OAB}$ with end poinst of $\vec{a}$ as $A$ and $\vec{b}$ as $B$. Angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$. Also $|\vec{a}|=|\vec{b}|$. So the other two angles are $75^{\circ}$. Can you now use geometry to find angle between $\vec{b}-\vec{a}$ and $x$ axis?