1989 - WAEC Mathematics Past Questions and Answers - page 2

L = 14cm, r = 7cm; \(\theta\)= ?

L = \(\frac{\theta}{360}\) x 2πr

14= \(\frac{\theta\}{360}\) x 2π(7)

\(\theta = \frac{360 \times 14}{44}\)

=114.55°

Let the integer = y

\(\therefore\) 5y - 2y = 63

3y = 63 \(\implies\) y = 21

3\(^y\) = 243;

3\(^y\) = 3\(^5\)

y = 5

tan\(\theta\) = 9/20 = 0.45

\(\theta\) = tan\(^{-1}\) 0.45 = 24.2°

the bearing of y from x = 180 0 - 0 = 180° - 24.2°

the bearing = 155.8°

\(\approxeq\) 156°

In \(\Delta DOB\), let < DOB = \(\alpha\)

In \(\Delta DOB\), \(5^2 = 3^2 + s^2\)

\(s^2 = 25 - 9 = 16\)

\(s = 4cm\)

\(\sin \alpha = \frac{4}{5}\)

\(\alpha = \frac{< AOB}{2}\)

Length of chord = \(2r \sin (\frac{\theta}{2})\)

|OB| = r = 5cm

L = \(2(5)(\frac{4}{5})\)

= 8 cm