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Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Solution
The interior angle of a regular polygon is twice the exterior angle. How many sides has the polygon?
A
5
B
6
C
8
D
9
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let the exterior angle = d
Note: Exterior angle + Interior angle = 180^\circ
\(\implies\) d + 2d = 180^\circ
3d = 180^\circ \(\implies\) d = 60^\circ
Recall, exterior angle = \(\frac{360}{\text{no of sides}}\)
\(\therefore \text{No of sides} = \frac{360}{60}\)
= 6 sides
Source: Myschool Past Questions Archive (Record #26387)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Solution
Which of the following figures have one line of symmetry only? I. Isosceles triangle II. Rhombus III. Kite
A
I and II only
B
II and III only
C
I and III only
D
I, II, and III
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Isosceles triangle and Kite shapes have 1 line of symmetry each while the rhombus has 2 lines of symmetry.
Source: Myschool Past Questions Archive (Record #26388)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
Calculate the value of Y in the diagram
Attached Figure / Diagram
A
18o
B
42o
C
48o
D
132o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x = 180^\circ - 123^\circ = 57^\circ
z = 180^\circ - 105^\circ = 75^\circ
y = 180^\circ - 57^\circ - 75^\circ = 48^\circ
Source: Myschool Past Questions Archive (Record #26391)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, POR is a circle with center O. ∠QPR = 50^\circ, ∠PQO = 30^\circ and ∠ORP = m. Find m.
Attached Figure / Diagram
A
20o
B
25o
C
30o
D
50o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
< QOR = 50^\circ x 2 = 100^\circ
Reflex < QOR = 360^\circ - 100^\circ = 260^\circ
\(\therefore\) 30^\circ + 50^\circ + 260^\circ + m = (4 - 2) x 180^\circ
340^\circ + m = 360^\circ
m = 360^\circ - 340^\circ = 20^\circ
Source: Myschool Past Questions Archive (Record #26392)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Solution
Given that m = -3 and n = 2 find the value of \(\frac{3n^2 - 2m^3}{m}\)
A
-22
B
-14
C
14
D
22
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Given that m = -3, n = 2, the value of
\(\frac{3n^2 - 2m^3}{m}\)
\(\frac{3(2)^2 -2(-3)^3}{-3} = \frac{12+54}{-3}\) = -22
Source: Myschool Past Questions Archive (Record #26393)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\frac{2-18m^2}{1+3m}\)
A
\(2(1+3m)\)
B
\(2(1+3m^2)\)
C
\(2(1-3m)\)
D
\(2(1-3m^2)\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{2-18m^2}{1+3m}=\frac{2(1-(3m)^2)}{1+3m}\\
\frac{2(1-(3m)(1+3m)}{1+3m}=2(1-3m)\)
Source: Myschool Past Questions Archive (Record #26394)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Solution
If \(y = \sqrt{ax-b}\) express x in terms of y, a and b
A
\(x = \frac{y^2-b}{a}\)
B
\(x = \frac{y+b}{a}\)
C
\(x = \frac{y-b}{a}\)
D
\(x = \frac{y^2 + b}{a}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(y = \sqrt{ax-b}\)
\(y^2 = ax-b\)
\(y^2 +b = ax\)
\(x = \frac{y^2 + b}{a}\)
Source: Myschool Past Questions Archive (Record #26395)
Mathematics
WAEC / WASSCE 1999
Indices, Logarithms & Surds
Solution
Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)
A
\(4.5\times 2^{n}\)
B
\(4.5\times 2^{2n}\)
C
\(9\times 2^{n-1}\)
D
\(9\times 2^{2n}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(K = 81 \times 2^{2n - 2}\)
\(\sqrt{K} = \sqrt{81 \times 2^{2n - 2}}\)
= \(9 \times 2^{n - 1}\)
Source: Myschool Past Questions Archive (Record #26396)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\frac{4}{x+1}-\frac{3}{x-1}\)
A
\(\frac{x+7}{x^2 - 1}\)
B
\(\frac{x-7}{x^2 + 1}\)
C
\(\frac{x-7}{x^2 - 1}\)
D
\(\frac{x-11}{x^2 - 1}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{4}{x+1}-\frac{3}{x-1} \\
=\frac{4x - 4 - 3x - 3}{(x+1)(x-1)}=\frac{x-7}{x^2 - 1}\)
Source: Myschool Past Questions Archive (Record #26397)
Mathematics
WAEC / WASSCE 1999
Indices, Logarithms & Surds
Solution
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)
A
\(\frac{\sqrt{3}+1}{2}\)
B
\(\frac{2}{\sqrt{3}+1}\)
C
\(\frac{\sqrt{3}-1}{2}\)
D
\(\frac{2}{\sqrt{3}-1}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)
Source: Myschool Past Questions Archive (Record #26399)
Mathematics
WAEC / WASSCE 1999
Trigonometry, Elevation & Bearings
Solution
From the top of a cliff 20m high, a boat can be sighted at sea 75m from the foot of the cliff. Calculate the angle of depression of the boat from the top of the cliff
A
14.9 o
B
15.5 o
C
74.5 o
D
75.1 o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
From the diagram above:
\(\tan \theta = \frac{20}{75} = 0.267\)
\(\theta = \tan^{-1} 0.267 = 14.93^\circ\)
\(\approxeq\) 14.9^\circ
Source: Myschool Past Questions Archive (Record #26400)
Mathematics
WAEC / WASSCE 1999
Trigonometry, Elevation & Bearings
Diagram
Solution
From the diagram above. ABC is a triangle inscribed in a circle center O. ∠ACB = 40º and |AB| = x cm. Calculate the radius of the circle.
Attached Figure / Diagram
A
\(\frac{x}{sin 40^o}\)
B
\(\frac{x}{cos 40^o}\)
C
\(\frac{x}{2 sin 40^o}\)
D
\(\frac{x}{2 cos 40^o}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the diagram given in the question, < ABC = 90º ( < in a semi-circle)
In triangle ABC, Sin 40º = \(\frac{x}{|AC|}\)
|AC| = \(\frac{x}{sin40º}\)
Since |AC| is a diameter, radius is half of |AC| = \(\frac{x}{2sin40º}\)
Source: Myschool Past Questions Archive (Record #26401)
Mathematics
WAEC / WASSCE 1999
Trigonometry, Elevation & Bearings
Diagram
Solution
The diagram shows the position of three ships A, B and C at sea. B is due north of C such that |AB| = |BC| and the bearing of B from A = 040^\circ. What is the bearing of A from C
Attached Figure / Diagram
A
040o
B
070o
C
110o
D
290o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
< ABC = 40^\circ (alternate angles)
\(\therefore\) < ACB = \(\frac{180^\circ - 40^\circ}{2}\)
= 70^\circ
\(\therefore\) Bearing of A from C = 360^\circ - 70^\circ
= 290^\circ
Source: Myschool Past Questions Archive (Record #26402)
Mathematics
WAEC / WASSCE 1999
Indices, Logarithms & Surds
Solution
A pole of length L leans against a vertical wall so that it makes an angle of 60o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L.
A
\(16\sqrt{3}m\)
B
\(4\sqrt{3}m\)
C
\(\frac{\sqrt{3}}{16}\)
D
\(\frac{16\sqrt{3}}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(sin 60 = \frac{8}{L} = 8 \div \frac{\sqrt{3}}{2}; L = \frac{8}{sin 60}=\frac{16}{\sqrt{3}}\)
When we rationalize gives \(\frac{16\sqrt{3}}{3}\)
Source: Myschool Past Questions Archive (Record #26403)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST.
Attached Figure / Diagram
A
60cm2
B
50cm2
C
40cm2
D
30cm2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
PR = 4 + 4 = 8 cm
Area of trapezium = \(\frac{1}{2} (a + b) h\)
= \(\frac{1}{2} (4 + 8) \times 5\)
= 30 cm\(^2\)
Source: Myschool Past Questions Archive (Record #26404)