1992 - WAEC Mathematics Past Questions and Answers - page 3

Draw line ANB// ML

< MNA = 40° (alternate angle AB/ ML)

< ANP = 55° - 40° = 15°

< POR = 15° = < ANP (corresponding angles, PN // RQ)

Sum of exterior angles = 360°

2x + 3x + 2x + 3x + 5x = 360°

15x = 360°

x = 360°/15 = 24°

< PTQ = 25° (base angles of an isos. triangle)

\(\therefore\) < PQT = 180° - (25° + 25°) = 130° (sum of angles in triangle PQT)

\(\therefore\) < RST = 130° - 75° = 55° (exterior angle = sum of 2 opp. interior angles)

In the second quadrant, \(\sin 120 = \sin (180 - 120)\)

= \(\sin 60\)

\(\cos \theta = \frac{5}{13}\)

\(\implies\) In the right- angled triangle, with an angle \(\theta\), the adjacent side to \(\theta\) = 5 and the hypotenuse = 13.

\(\therefore 13^2 = opp^2 + 5^2\)

\(opp^2 = 169 - 25 = 144 \implies opp = \sqrt{144}\)

= 12.

\(\tan \theta = \frac{opp}{adj} = \frac{12}{5}\) 

tan 315° = - tan (360 - 315) = - tan 45

= -1

sin 210 = - sin (210 - 180) = - sin 30

= \(-\frac{1}{2}\)