Find the angle between forces of magnitude 7N and 4N if their resultant has a magnitude of 9N.

  • A 39.45°
  • B 73.40°
  • C 75.34°
  • D 106.60°

The correct answer is B. 73.40°

\(F_{1} = 7i + 0j\)

\(F_{2} = (4\cos\theta)i + (4\sin\theta)j\)

\(9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}\)

\(9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta\)

\(81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)\)

Recall, \(\cos^{2}\theta + \sin^{2}\theta = 1\)

\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)

\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)

\(\theta = \cos^{-1} 0.2857  = 73.40°\)

Previous question Next question