Question

Derivation of tan alpha= b sin theta/a+b cos theta

Answer 1

Answer:

The direction of the resultant vector is

$tan\alpha=\dfrac{bsin\theta}{a+bcos\theta}$

Step-by-step explanation:

Consider two vectors $\vec A$ and $\vec B$ as shown in the figure.

According to the triangle law of vectors, the resultant of the sum of the two vectors is given by the third side of the triangle. i.e.,

$\vec R=\vec A+\vec B$  

Drop a perpendicular from B onto vector $\vec A$ to point N and extend $\vec A$ to join N.

Thus, ABN is a right angled triangle. And

$cos\theta=\dfrac{adj}{hyp} =\dfrac{AN}{\vec B}\implies \vec Bcos\theta=AN$

$sin\theta=\dfrac{opp}{hyp} =\dfrac{BN}{\vec B}\implies \vec Bsin\theta=BN$

Applying the Pythagorean theorem to the triangle AON,

$OB^{2}= ON^{2} +BN^{2}$

$\implies R^{2}= (OA+AN)^{2} +BN^{2}$

            $= (a+bcos\theta)^{2} +(bsin\theta)^{2}$

            $= a^{2}+b^{2}cos^{2}\theta+2abcos\theta +b^{2}sin^{2}\theta$

            $= a^{2}+b^{2}(cos^{2}\theta+sin^{2}\theta)+2abcos\theta$

            $= a^{2}+b^{2}+2abcos\theta$

Therefore,  the magnitude of the resultant vector is

$R= \sqrt{a^{2}+b^{2}+2abcos\theta }$

In the triangle AON, if the angle between vector $\vec A$ and the resultant vector $\vec R$ is $\alpha$, then

$tan\alpha = \dfrac{opp}{adj}$

       $= \dfrac{BN}{AN} =\dfrac{BN}{OA+AN}$

      $=\dfrac{bsin\theta}{a+bcos\theta}$  

Therefore, the direction of the resultant vector is given by

$tan\alpha=\dfrac{bsin\theta}{a+bcos\theta}$

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