Let \[\vec{A} \text { and } \vec{B}\] be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angle 30° and 60° respectively, find the resultant.

#### Solution

Angle between \[\vec{A} \text { and } \vec{B}\],θ = 60° − 30° = 30°

\[\left| \vec{A} \right| = \left| \vec{B} \right| = 10 \text { units }\]

The magnitude of the resultant vector is given by

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

\[ = \sqrt{{10}^2 + {10}^2 + 2 \times 10 \times 10 \times \cos 30^\circ}\]

\[ = \sqrt{200 + 200 \cos 30^\circ}\]

\[ = \sqrt{200 + 100}\]

\[ = \sqrt{300} = 17 . 3 \text { units }\]

Let β be the angle between

\[\vec{R} \text { and } \vec{A}\].

\[\therefore \beta = \tan^{- 1} \left( \frac{A \sin 30^\circ}{A + A \cos 30^\circ} \right)\]

\[ \Rightarrow \beta = \tan^{- 1} \left( \frac{10 \sin 30^\circ}{10 + 10 \cos 30^\circ} \right)\]

\[ \Rightarrow \beta = \tan^{- 1} \left( \frac{1}{2 + \sqrt{3}} \right) = \tan^{- 1} \left( \frac{1}{3 . 372} \right)\]

\[ \Rightarrow \beta = \tan^{- 1} \left( 0 . 26795 \right) = 15^\circ\]

Angle made by the resultant vector with the X-axis = 15° + 30° = 45°

∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.