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Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
From the diagram, find the bearing of Q from P.
Attached Figure / Diagram
A
236o
B
214o
C
146o
D
124o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #26536)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram, SQ is a tangent to the circle at P, XP||YQ, ∠XPY = 56 o and ∠PXY = 80 o.Find angle PQY
Attached Figure / Diagram
A
34o
B
13.36o
C
44o
D
46o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
< XYQ = 180^\circ - (80^\circ + 56^\circ)
= 44^\circ
< PYQ = 56^\circ (alternate angles, XP||YQ)
< QPY = 90^\circ
< PQY = 180^\circ - (90^\circ + 56^\circ)
= 34^\circ
Source: Myschool Past Questions Archive (Record #26540)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Solution
The lengths of the parallel sides of a trapezium are 9 cm and 12 cm. lf the area of the trapezium is 105 cm\(^2\), find the perpendicular distance between the parallel sides.
A
5cm
B
7cm
C
10cm
D
15cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{1}{2}(a+b)\times h = 105cm^2\)
\(\frac{1}{2}(9+12)\times h = 105\)
h = \(\frac{105 \times 2}{21} = 10cm\)
Source: Myschool Past Questions Archive (Record #26543)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram O and O' are the centres of the circles radii 15cm and 8cm respectively. If PQ = 12cm, find |OO'|.
Attached Figure / Diagram
A
8.46cm
B
19.04cm
C
20.81cm
D
26.16cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In \(\Delta\) POQ,
\(12^2 = 15^2 + 15^2 - 2(15)(15) \cos < POQ\)
\(144 = 450 - 450\cos < POQ\)
\(450 \cos < POQ = 450 - 144 = 306\)
\(\cos
Source: Myschool Past Questions Archive (Record #26549)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram O is the centre of the circle. Which of the following is/are not true? I. a = b II. b + c = 180o III. a + b = c
Attached Figure / Diagram
A
I and II only
B
II and III only
C
II only
D
III only
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
i.a and b are 90( angles in same segment are equal)
ii. angle b+c=180 (opposite angles in a cyclic quadrilateral)
iii. a=90 and b=90 .:a+b=180 which is not equal to c
Source: Myschool Past Questions Archive (Record #26550)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Solution
The angle of depression of a point Q from a vertical tower PR, 30m high, is 40^\circ. If the foot P of the tower is on the same horizontal level as Q, find, correct to 2 decimal places, |PQ|.
A
35.75m
B
25.00m
C
22.98m
D
19.28
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\tan 40^\circ = \frac{30}{|PQ|}\)
\(|PQ| = \frac{30}{\tan 40^\circ} = \frac{30}{0.839}\)
= 35.75m
Source: Myschool Past Questions Archive (Record #26553)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram, QRT is a straight line. If angle PTR = 90^\circ, angle PRT = 60^\circ, angle PQR = 30^\circ and |PQ| = \(6\sqrt{3}cm\), calculate |RT|
Attached Figure / Diagram
A
0.3cm
B
\(\frac{\sqrt{3}}{2}cm\)
C
3cm
D
\(3\sqrt{3}cm\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In \(\Delta\) QPT,
\(\frac{PT}{6\sqrt{3}} = \sin 30^\circ\)
PT = \(6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3} cm\)
In \(\Delta\) RPT,
\(\frac{PT}{RT} = \tan 60^\circ\)
\(\frac{3\sqrt{3}}{RT} = \tan 60^\circ\)
\(RT = \frac{3\sqrt{3}}{\sqrt{3}} = 3 cm\)
Source: Myschool Past Questions Archive (Record #26558)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram, calculate the value of x
Attached Figure / Diagram
A
35o
B
80o
C
100o
D
115o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
From the diagram above.
115º = A (corresponding angles)
t + 115º = 180^\circ ( sum of
Source: Myschool Past Questions Archive (Record #26559)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30^\circ and angle PQR = 45^\circ. Calculate in meters |QS|
Attached Figure / Diagram
A
\(10(1+\sqrt{3})\)
B
\(20\sqrt{3}\)
C
\(10\sqrt{3}\)
D
\((10+\sqrt{3})\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In \(\Delta\) PQR,
\(\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45\)
= 10m
In \(\Delta\) PRS,
\(\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}\)
= \(\frac{10}{\frac{1}{\sqrt{3}}\)
= \(10\sqrt{3}\)
PS = \(10 + 10\sqrt{3}\)
= \(10(1 + \sqrt{3}) cm\)
Source: Myschool Past Questions Archive (Record #26561)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Solution
To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?
A
60km/hr
B
80km/hr
C
90km/hr
D
108km/hr
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(speed = \frac{distance}{time}\\
72 = \frac{60}{time}\\
t = \frac{60}{72} = \frac{5}{6}hr\)
time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\)
Time required for the journey
\(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\
speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)
Source: Myschool Past Questions Archive (Record #26563)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram, \(QR||TP and W\hat{P}T = 88^{\circ} \). Find the value of x
Attached Figure / Diagram
A
92o
B
68o
C
67o
D
23o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Sum of the angles in a triangle = 180^\circ
3x - 180^\circ + 92^\circ + x = 180^\circ
4x - 88^\circ = 180^\circ
4x = 268^\circ
x = 67^\circ
Source: Myschool Past Questions Archive (Record #26564)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram O is the center of the circle, ∠SOR = 64^\circ and ∠PSO = 36^\circ. Calculate ∠PQR
Attached Figure / Diagram
A
100o
B
96o
C
94o
D
86o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
< OSR = < ORS = \(\frac{180^\circ - 64^\circ}{2}\) = 58^\circ
< PSR = 36^\circ + 58^\circ = 94^\circ
< PSR + < PQR = 180^\circ
94^\circ + < PQR = 180^\circ \(\implies\) < PQR = 180^\circ - 94^\circ = 86^\circ
Source: Myschool Past Questions Archive (Record #26565)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram, LMT is a straight line. lf O is the centre of circle LMN, OMN = 20^\circ, LTN = 32^\circ and |NM| = |MT|, find LNM.
Attached Figure / Diagram
A
44o
B
46o
C
52o
D
70o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
< MNT = < MTN = 32^\circ
< NMT = 180^\circ - 2(32^\circ) = 116^\circ
< OMN + < NMT + < LMO = 180^\circ
20^\circ + 116^\circ + < LMO = 180^\circ \(\implies\) < LMO = 180^\circ - 136^\circ = 44^\circ
< LMN = < LMO + < OMN
= 44^\circ + 20^\circ = 64^\circ
< NOM = 180^\circ - 2(20^\circ) = 140^\circ
< NLM = \(\frac{1}{2} \times < NOM = 70^\circ\)
< LNM + < LMN + < NLM = 180^\circ
< LNM + 64^\circ + 70^\circ = 180^\circ
< LNM = 180^\circ - 134^\circ = 46^\circ
Source: Myschool Past Questions Archive (Record #26569)
Mathematics
NECO SSCE 2002
Geometry and Trigonometry
Solution
Evaluate 1.25 \times 10\(^2\) \times 0.125 \times 10\(^3\) \times 12.5 \times 10\(^{-4}\) leaving your answer in standard form
A
1.995 × 103
B
1.95 × 102
C
1.95 × 101
D
1.95 × 10-2
E
1.95 × 10-3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Where 1.25 \times 10\(^2\) = 125
0.125 \times 10\(^3\) = 125
12.5 \times 10\(^{-4}\) = 0.00125
: 125 x 125 x 0.00125
= 19.53125 or 1.95 x 10\(^1\)
Source: Myschool Past Questions Archive (Record #42011)
Mathematics
NECO SSCE 2002
Geometry and Trigonometry
A cone was developed from a sector of a circle with radius10cm and an angle at the centre of 108°. What is the radius of the cone?
A
9.0cm
B
8.5cm
C
7.2cm
D
4.3cm
E
3.0cm
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #42021)