# The resultant of the two forces 3N and2N at an angle theta is doubled when first force in increased to 6N. Find theta?

Mar 7, 2016

$\theta = \frac{2 \pi}{3}$ or ${120}^{\circ}$

#### Explanation:

Formula for resultant force of two forces $P$ and $Q$ is given by $R = \sqrt{{P}^{2} + {Q}^{2} + 2 P \times Q \times \cos \theta}$.

When two forces $3 N$ and $2 N$ are at an angle $\theta$, value of the first resultant is ${R}_{1}$ is given by

${R}_{1} = N \sqrt{{3}^{2} + {2}^{2} + 2 \times 3 \times 2 \times \cos \theta} = \sqrt{13 + 12 \cos \theta}$

When first force is increased to $6 N$, the resultant force ${R}_{2}$ is given by

${R}_{2} = N \sqrt{{6}^{2} + {2}^{2} + 2 \times 6 \times 2 \times \cos \theta} = \sqrt{40 + 24 \cos \theta}$

As ${R}_{2}$ is doubled

$N \sqrt{40 + 24 \cos \theta} = 2 \times N \sqrt{13 + 12 \cos \theta}$ or

$40 + 24 \cos \theta = 4 \times \left(13 + 12 \cos \theta\right) = 52 + 48 \cos \theta$

Hence, $24 \cos \theta = - 12$ or $\cos \theta = - \frac{1}{2}$

and $\theta = \frac{2 \pi}{3}$ or ${120}^{\circ}$