A vector \[\vec{A}\] makes an angle of 20° and \[\vec{B}\] makes an angle of 110° with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.

#### Solution

From the above figure, we have:

Angle between \[\vec{A}\] and \[\vec{B}\] = 110° − 20° = 90° \[\left| \vec{A} \right| = 3 \text{ m } \text{ and } \left| \vec{B} \right| = \text{ 4 }m\]

Magnitude of the resultant vector is given by

\[R = \sqrt{A^2 + B^2 + 2AB \cos \theta}\]

\[ = \sqrt{3^2 + 4^2 + 2 \times 3 \times 4 \times \cos 90^\circ}\]

\[ = \text{ 5 }m\]

Let β be the angle between \[\vec{R} \text { and } \vec{A}\].

\[\beta = \tan^{- 1} \left( \frac{A \sin 90^\circ}{A + B \cos 90^\circ} \right)\]

\[ = \tan^{- 1} \left( \frac{4 \sin 90^\circ}{3 + 4 \cos 90^\circ} \right)\]

\[ = \tan^{- 1} \frac{4}{3}\]

\[ = \tan^{- 1} \left( 1 . 333 \right)\]

\[ = 53^\circ\]

Now, angle made by the resultant vector with the X-axis = 53° + 20° = 73°

∴ The resultant \[\vec{R}\] is 5 m and it makes an angle of 73° with the x-axis.