WAEC Further Mathematics 2015 Paper

$\overset{\rightarrow}{x}. \overset{\rightarrow}{y} = \left|\overset{\rightarrow}{x}\right|\left|\overset{\rightarrow}{y}\right|\cos\theta$

$\overset{\rightarrow}{x}. \overset{\rightarrow}{y} = (i - 3j) . (6i + j) = 6 - 3 = 3$

$\left|\overset{\rightarrow}{x}\right| = \sqrt{1^{2} + (-3)^{2}} = \sqrt{10}$

$\left|\overset{\rightarrow}{y}\right| = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$

$\therefore 3 = (\sqrt{10})(\sqrt{37})\cos\theta$

$\cos\theta = \frac{3}{\sqrt{370}} = 0.1559$

$\theta = \cos^{-1} 0.1559 \approx 81^{\circ}$