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Question
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.
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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(100 + 100 + 200 + sqrt3/2)`
= `sqrt(200 +100 xx sqrt3)`
= `sqrt(200 + 173)`
= `sqrt373`
= 19.3
Let β be the angle between
\[\vec{R} \text { and } \vec{A}\].
\[\therefore \beta = \tan^{- 1} \left( \frac{A \sin 30^\circ}{A + A \cos 30^\circ} \right)\]
\[ \Rightarrow \beta = \tan^{- 1} \left( \frac{10 \sin 30^\circ}{10 + 10 \cos 30^\circ} \right)\]
`beta = tan^(-1) (10 xx 1/2)/(10 + 10 xx (sqrt3)/2)`
`beta = tan^(-1) 5/(5(2 + sqrt3))`
`beta = tan^(-1) (1/(2 + sqrt3))`
\[ \Rightarrow \beta = \tan^{- 1} = 15^\circ\]
Angle made by the resultant vector with the X-axis = 15° + 30° = 45°
∴ The magnitude of the resultant vector is 19.3, and it makes an angle of 45° with the X-axis.
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