2006 - WAEC Mathematics Past Questions and Answers - page 4

Temperature = -5^oC

It fell by 6^oC = -5^oC - 6^oC

= -11^oC

Then rose by 7^oC = -11 + 7

= -4^oC (final temperature)

\(\frac{20}{5 \sqrt{4 \times 7} - 2\sqrt{9 \times 7}}\)

\(\frac{20}{10 \sqrt{7} - 6\sqrt{7}}\)

= \(\frac{20 \sqrt{7}}{4 \sqrt{7}}\)

= \(\frac{5\sqrt{7}}{7}\)

Sin 65^o = \(\frac{x}{16}\)

x = 16x sin 65

= 16 x 0.9063

x = 14.5m

|VU| = |UT| = |TS|

|VS| = (4.8) x 3 = 14.4cm

|PQ| = |QR| = 4.8

|PR| = (4.8) x 2 = 9.6cm

Since quad. PRSV is a trapezium of height 5cm

Area of quad. PRSV = \(\frac{1}{2}(a + b)h\)

= \(\frac{1}{2}(14.4 + 9.6) \times 5\)

= \(\frac{1}{2}(24) \times 5\)

12 x 5 = 60cm²

tan 6^o = \(\frac{|PR|}{|QR|}\)

\(\sqrt{3} = \frac{|PR|}{5}\)

= |PR| = \(5 \sqrt{3}\)cm

sin 45 = \(\frac{|PR|}{|PS|}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{5 \sqrt{3}}{|PS|}\)

|PS| = \(5 \sqrt{3}\) x \(\sqrt{2}\)

= 5\(\sqrt{6}\)cm

PQR = QRS = Y(alt. amgles)

PQR + x = 180^o(angles on a straight line)

y + x = 180^o

PRQ = 90 - 58 = 32^o(angle in a semi-circle)

Since PRS = 90^o(radius angular to tangent)

QRS = 90 + 32

= 122^o

Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50^o

Also, < QSR = < QPR(angles in the segment)

< QPR = x^o

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130^o

Q₁ = 50(alternate angle)

Q₂ = 180^o - 110^o (straight line angle)

Q₂ = 70

PQR = Q₁ + Q₂

= 50^o + 70^o

= 120^o

area of larger circle = 100\(\pi\)

\(\pi r^2 = 100\pi\)

R² = 100

R = 10(Radius)

Area of smaller circle = \(49 \pi\)

\(\pi r^2 = 49\pi\)

r² = 49

r = 7(radius)

Since R = x + r

x = R - r

x = 10 - 7 = 3