1995 - WAEC Mathematics Past Questions and Answers - page 4

< DBC = 35° (base angles of an isosceles triangle)

< CDB = 180° - (35° + 35°)

= 110°

< ADB = 70°; < ADB = 90°

\(\therefore\) < BAO = 180° - (70° + 90°)

= 20°

4x + 3x + 2x + 90 =360° [angle at a point]

9x + 90 = 360°

9x = 360° - 90°

9x = 270

x = 270/9

x = 30°

y = 86° - 50° [Alternate ∠s are equal]

y = 86° - 50° = 36°

a = y = 36° [Alternate ∠s]; b = 64° [Alternate ∠s;]

x = a + b = (36° + 64°)

=100°

log\(_{10}\) x = \(\bar{2}.3675\) => x = 0.023

log\(_{10}\) y = \(\bar{2}.9738\) => y = 0.094

x + y = 0.117

Ae is a bisector of BAC => EAC = 45^o
CE is the bisector of ACD => ACE = 45^o
AEC = 180^o - [EAC + ACE]
    = 180^o - (45^o + 45^o) = 180^o - 90^o = 90^o

a = 52^o [vertical opp. ∠s are equal]
a = b = 52^o [Alternate ∠s]
b = c = 52^o [vertical opp ∠s]
c + x + 35^o = 180^o [sum of angles in a triangle]
52^o + x + 35^o = 180^o
x + 87^o = 180^o
x = 180^o - 87^o = 93^o

a + c + d - 180^o[sum of angles in triangle]
x = a[vertical opp. ∠s]
y + x + d = 180^o[sum of ∠s in a Δ]
y + a + d = 180^o
y + a + d = a + c + d
y = c
but b + y = 180^o [sum of ∠s on a straight line ]
b + c = 180^o; Ans = A = 180^o

The formula for the exterior angle of a regular polygon = \(\frac{360}{n}\).

Among the options, only 72° is divisible by 360° to give an integer.

ACB = ABC [Base angles of isoceles triangle]
ACB + ABC + 50^o = 180^o[sum of angles in a triangle]
ACB + ABC = 180^o - 50^o = 130^o