Vectors vec{A} and vec{B} make angles of 20^c | vedclass

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(B) The angle between vectors $\vec{A}$ and $\vec{B}$ is $	heta = 110^\circ - 20^\circ = 90^\circ$.The magnitude of the resultant vector $\vec{R}$ is

Vedclass · Accepted Answer

$	an^{-1}(12/5) + 20^\circ$

Vedclass · Answer

(B) The angle between vectors $\vec{A}$ and $\vec{B}$ is $	heta = 110^\circ - 20^\circ = 90^\circ$.The magnitude of the resultant vector $\vec{R}$ is given by:$R = \sqrt{A^2 + B^2 + 2AB \cos(90^\circ)} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \ m$.Let $\alpha$ be the angle that the resultant vector $\vec{R}$ makes with vector $\vec{A}$.Using the formula for the direction of the resultant:$	an \alpha = \frac{B \sin 	heta}{A + B \cos 	heta} = \frac{12 \sin 90^\circ}{5 + 12 \cos 90^\circ} = \frac{12 	imes 1}{5 + 12 	imes 0} = \frac{12}{5}$.Therefore,$\alpha = 	an^{-1}(12/5)$.Since vector $\vec{A}$ itself makes an angle of $20^\circ$ with the $x$-axis,the angle of the resultant vector $\vec{R}$ with the $x$-axis is $\alpha + 20^\circ = 	an^{-1}(12/5) + 20^\circ$.

Vectors $\vec{A}$ and $\vec{B}$ make angles of $20^\circ$ and $110^\circ$ with the $x$-axis,respectively. The magnitudes of these vectors are $5 \ m$ and $12 \ m$,respectively. The angle that their resultant vector makes with the $x$-axis is:

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