Physics, asked by Anonymous, 1 year ago

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Don't copy from internet. Paap lagega! I need the written solution!

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Answered by Anonymous

hope it helps u...............

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Answered by Anonymous

Given :

▪ Two vectors having equal magnitudes A make an angle $\theta$ with each other.

To Find :

▪ Magnitude and direction of the resultant vector.

Concept :

☞ By triangle law or parallologram law of vector addition, the magnitude of resultant R at two vectors P and Q inclined to each other at angle $\theta$ , is given by

$\bigstar\:\underline{\boxed{\bf{\pink{R=\sqrt{P^2+Q^2+2PQ\cos\theta}}}}}$

☞ If resultant R makes an angle β with P, then

$\bigstar\:\underline{\boxed{\bf{\green{\tan\beta=\dfrac{Q\sin\theta}{P+Q\cos\theta}}}}}$

Calculation :

☢ Magnitude of resultant vector :

$\dashrightarrow\sf\:R=\sqrt{P^2+Q^2+2PQ\cos\theta}\\ \\ \dashrightarrow\sf\:R=\sqrt{A^2+A^2+2A^2\cos\theta}\\ \\ \dashrightarrow\sf\:R=\sqrt{2A^2+2A^2\cos\theta}\\ \\ \dashrightarrow\sf\:R=\sqrt{2A^2(1+\cos\theta)}\\ \\ \dag\sf\:\:\gray{1+\cos\theta=2\sin^2\dfrac{\theta}{2}}\\ \\ \dashrightarrow\sf\:R=\sqrt{4A^2\sin^2\dfrac{\theta}{2}}\\ \\ \dashrightarrow\underline{\boxed{\bf{\red{R=2A\sin\dfrac{\theta}{2}}}}}\:\orange{\bigstar}$

☢ Direction of resultant vector :

$\implies\sf\:\tan\beta=\dfrac{Q\sin\theta}{P+Q\cos\theta}\\ \\ \implies\:\tan\beta=\dfrac{A\sin\theta}{A(1+\cos\theta)}\\ \\ \implies\sf\:\tan\beta=\dfrac{\sin\theta}{2\cos^2\frac{\theta}{2}}\\ \\ \implies\sf\:\tan\beta=\dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\\ \\ \implies\sf\:\tan\beta=\dfrac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\ \\ \implies\sf\:\tan\beta=\tan\frac{\theta}{2}\\ \\ \implies\underline{\boxed{\bf{\blue{\beta=\dfrac{\theta}{2}}}}}\:\orange{\bigstar}$

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Given :

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Don't copy from internet. Paap lagega! I need the written solution!