(a) --- (iv) because sin θ = 0º , hence | A | | B | sin0 = 2×4×0 = 0
(b) --- (iii) because sin 90º = 1, hence | A | | B | sin 90º = 2×4×1 = 8
(c) --- (i) because sin 30º = \(\frac{1}{2}\) , hence | A | | B | sin 30º = 2×4× \(\frac{1}{2}\) = 4
(d) --- (ii) because sin 45º = \(\frac{1}{\sqrt 2}= \frac{\sqrt 2}{2}\) , hence, | A | | B | sin 45º = 2×4×\(\frac{\sqrt 2}{2}=4\sqrt2\)