WAEC Mathematics Past Questions & Answers - Page 90

tan \(\theta = \frac{opp}{adj} = \frac{6.6}{13}\)

tan \(\theta = 0.5077\)

\(\theta\) = tan^-1 0.5077

\(\theta = 27^o\)

\(\frac{1}{2}\)(a - b + c) + \(\frac{1}{2}\)(a + b - c) - [\(\frac{1}{2}\) (a - b - c)]

\(\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)

= \(\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)

= \(\frac{1}{2}(a + b + c)\)

\(\frac{2}{x - 3} - \frac{3}{x - 2}\) = \(\frac{p}{(x - 3)(x - 2)}\)

\(\frac{2(x - 2) - 3(x - 3)}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\)

= \(\frac{2x - 4 - 3x + 9}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\)

\(\frac{-x + 5}{(x - 3)(x - 2)} = \frac{p}{(x - 3)(x - 2)}\)

p(x - 3)(x - 2) = -x + 5(x - 3)(x - 2)

p = -x + 5 or p = 5 - x

\(\frac{\sqrt{8^2 \times 4^{n + 1}}}{2^{2n} \times 16}\)

= \(\frac{\sqrt{2^{3(2)} \times 2^{2(n + 1)}}}{2^{2n} \times 2^4}\)

= \(\frac{\sqrt{2^6 \times 2^{2n + 2)}}}{2^{2n} + 4}\)

= \(\frac{\sqrt{2^6 + 2^{2n + 2)}}}{2^{2n} + 4}\)

= \(\frac{\sqrt{2^{2n + 8}}}{2^{2n} + 4}\)

= \(\sqrt{2^{2n + 8} \div 2^{2n} + 4}\)

= \(\sqrt{2^{2n - 2n} + 8 - 4}\)

= \(\sqrt{2^4}\)

= \(\sqrt{16}\)

= 4

No explanation has been provided for this answer.