Question

The resultant of two forces $P$ and $Q$ acting at an angle $\theta$ is equal to $(2m+1)\sqrt{P^2+Q^2}$, when they act at an angle $90-\theta$, the resultant is $(2m-1)\sqrt{P^2+Q^2}$; then $\tan \left(\theta \right)=$

• A. $\frac{1}{m}$
• B. $\frac{m+1}{m-1}$
• C. $\frac{m-1}{m+1}$
• D. $\sqrt{1+m^2}$

Answer 1

The correct option is C

$\frac{m-1}{m+1}$

The explanation for the correct option

Step 1: Given data

The resultant of $P$ and $Q$acting at an angle $\theta$ is $(2m+1)\sqrt{P^2+Q^2}$.

The resultant of $P$ and $Q$acting at an angle $90-\theta$ is $(2m-1)\sqrt{P^2+Q^2}$.

Step 2: Find $\cos \left(\theta \right)$ when the two forces $P$ and $Q$ acting at an angle $\theta$

$$\begin{gathered} \left|\overrightarrow{P}+\overrightarrow{Q}\right|=\sqrt{P^2+Q^2+2PQ\cos \left(\theta \right)} \\ \left(2m+1\right)\sqrt{P^2+Q^2}=\sqrt{P^2+Q^2+2PQ\cos \left(\theta \right)} \\ {\left(2m+1\right)}^2\left(P^2+Q^2\right)=P^2+Q^2+2PQ\cos \left(\theta \right) \\ \frac{4m^2+4m}{2PQ}\left(P^2+Q^2\right)=\cos \left(\theta \right)........\left(i\right) \end{gathered}$$

Step 3: Find $\cos \left(90°-\theta \right)$ when the two forces $P$ and $Q$ acting at an angle $90°-\theta$

$$\begin{gathered} \left|\overrightarrow{P}+\overrightarrow{Q}\right|=\sqrt{P^2+Q^2+2PQ\cos \left(\theta \right)} \\ \left(2m-1\right)\sqrt{P^2+Q^2}=\sqrt{P^2+Q^2+2PQ\cos \left(90°-\theta \right)}{\left(2m-1\right)}^2\left(P^2+Q^2\right)=P^2+Q^2+2PQ\sin \left(\theta \right) \\ \frac{4m^2-4m}{2PQ}\left(P^2+Q^2\right)=\sin \left(\theta \right).....\left(ii\right) \end{gathered}$$

Step 4: Find the value of $\tan \left(\theta \right)$

$$\begin{gathered} \therefore \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)} \\ =\frac{\frac{4m^2-4m}{2PQ}\left(P^2+Q^2\right)}{\frac{4m^2+4m}{2PQ}\left(P^2+Q^2\right)} \\ =\frac{m-1}{m+1} \end{gathered}$$

Hence option (C) is the correct answer.