Express the force F = (8 N, 150°) in the form (a i + b j) where a and b are constants
\(4\sqrt{3} i - 4j\)
\(4 i - 4\sqrt{3} j\)
\(- 4 i + 4\sqrt{3} j\)
\(- 4\sqrt{3} i + 4j\)
Correct answer is D
\(F = F\cos \theta i + F \sin \theta j\)
\((8 N, 150°) = 8 \cos 150 i + 8 \sin 150 j\)
= \(- 8 \cos 30 i + 8 \sin 30 j\)
= \(-8(\frac{\sqrt{3}}{2}) i + 8(\frac{1}{2} j\)
= \(-4\sqrt{3} i + 4 j\)