2003 - JAMB Mathematics Past Questions and Answers - page 3

Since PQ = QR, Δ PQR is a right angle isosceles triangle
∴ ∠PQR = 45^o
=> The bearing of P from R = 180 + 45
= 225^o

θ = 30 x 2(∠ at center twice ∠ at θ)
∴ θ = 60^o
Length of arc = \(\frac{\theta}{360}\times 2\pi r\\ =\frac{60}{360}\times 2 \times \frac{22}{7}\times \frac{21}{1}\\ =22cm\)

Locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1) is the perpendicular bisector of the straight line joining Q and R
Mid point QR = (x₂+x₁)/2 . (y₂+y₁)/2
= ²⁺⁰/₂ . ¹⁺⁰/₂

Gradient of Qr =	y2-y1
x2-x1

=	1-0
2-0

= ¹/₂

Gradient of PM(M)=	-1
1/2

= -2
Equation of Pm = y - y₁ = m(x-x₁)
i.e y - ¹/₂ = -2(x-1)
2y - 1 = -4(x-1)
2y - 1 = -4x + 4
2y + 4x = 5

γ + S = 180(Allied angles)
S = 180 - γ
α + P = 180(Allied angles)
p = 180 - α
β = P + S
β = 180 - α + 180 - γ
β = 360 - α - γ
β + α + γ = 360

θ = 45 x 2 (∠ at center twice ∠ at θ)
θ = 90
Area of sector = ^θ/₃₆₀ x πr²
= ⁹⁰/₃₆₀ x ²²/₇ x 7 x 7
= ⁷⁷/₂cm²
Area of triangle = ¹/₂ ab sin θ
= ¹/₂ x 7 x 7 sin 90
= ¹/₂ x 7 x 7 x 1
= ⁴⁹/₂
Area of shaded region = ⁷⁷/₂ - ⁴⁹/₂

1	1
2

= ²⁸/₂
= 14cm²

R = P (base ∠s of isc Δ, ps = RS)
∴R = 40^o
∠PSQ = R (∠s in alternate segment)
∴∠PSQ = 40^o

Midpoint of PQ \(\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\\ =\frac{-1+(-3)}{2};\frac{6-4}{2}\\ =\frac{-4}{2};\frac{2}{2}\\ =-2,1\)
Midpoint of QR = \(\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\\ \frac{-3+1}{2};\frac{-4-4}{2}\\ \frac{-2}{2};\frac{-8}{2}\\ -1,-4\)
Midpoint of PQ and QR = (-2,1) and (-1,-4)

PS = SR

\(\bigtriangleup\)PSR is an isosceles

< P = < SPQ = 40^o

< R = < SRQ

< PSQ = < SRQ (alternate segment)

< PSQ = 40^o