WAEC Mathematics Past Questions & Answers - Page 123

The bearing of x from y = 270^o + \(\theta\)

where \(\theta\) + 50^o = y

in \(\bigtriangleup\) OXY

O + X + Y = 180^o

Where O = 40^o + 30^o = 70^o

70^o + 30^o + y = 180^o

y + 100^o = 180^o

y = 180^o - 100^o = 30^o

\(\theta\) + 50^o = 80^o

80^o - 50^o = 30^o

The bearing of x from y = 270^o + 30^o = 300^o

POQ in a straight line

Hence, < POQ + < QOR = 180^o

56^o + < QOR = 180^o

< QOR = 180^o - 56^o

= 124^o

Now, in \(\bigtriangleup\) QOR OR = OQ = Radius

< ORQ = < OQR = 2x (Base angles of an Isosceles \(\bigtriangleup\))

2x + 124 + 2x = 180^o

4x + 124 = 180

4x = 180 - 124

4x = 56

x = \(\frac{56}{4}\)

x = 14^o

Area of remaining = Area of rectangle = Area of semi-circle

22 x 8 - \(\frac{1}{2}\)xar²

ehere r - \(\frac{14}{2}\)cm = 7cm

Area of remaining = 176 - \(\frac{1}{2}\) x \(\frac{22}{4}\) x 7 x 7

= 176 - 77

= 99cm²

Total surface area = sum of the area of the \(\bigtriangleup\) S + Area of the rectangle

= 2 x Area of \(\bigtriangleup\) PTQ + 2 x Area of \(\bigtriangleup\) QUR + Area of rectangle PQRS

2 x \(\frac{1}{2}\)(8 x 4) + 2 x \(\frac{1}{2}\)(5 x 4) + 8 x 5

= 32 + 20 + 40

= 92cm²

F = 54^o (Alternate triangle angles)

< GHE = F = 54^o(Angle between a chord and a tangent = angles in the alternate segment)

Now, < GHE + < EHF + < IHF = 180^o(Angles on a straight line)

54^o + < EHF + 54^o = 180^o

< EHF = 180^o - 108^o

= 72^o