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Mathematics
UTME / JAMB 1994
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQRS is a parallelogram. Find the value of < SQR
Attached Figure / Diagram
A
30o
B
50o
C
80o
D
100o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR
< STP = < RQV = 30o
But SQR + 30o + 50o = 180o
SQR = 180 - 80
= 100o
Source: Myschool Past Questions Archive (Record #33036)
Mathematics
UTME / JAMB 1994
Geometry and Trigonometry
Diagram
Solution
In the diagram, O is the centre of the circle. If SOQ is a diameter and o, what is the value of < PSQ
Attached Figure / Diagram
A
148o
B
104o
C
80o
D
52o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
o (angle in the same segment of a circle are equal)
But o (angle in a semicircle)
also o (angles in a triangle = 180o)
o + 90o = 190o
o = 180o
PSQ = 180o - 128o
PSQ = 52o
Source: Myschool Past Questions Archive (Record #33037)
Mathematics
UTME / JAMB 1994
Geometry and Trigonometry
Diagram
Solution
In the diagram, PTS is a tangent to the circle TQR at T. Calculate < RTS
Attached Figure / Diagram
A
120o
B
70o
C
60o
D
40o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180
RQT = 180o - (50 + 60)
= 180o - 110o
= 70o
Since RQT = RTS = 70o
Source: Myschool Past Questions Archive (Record #33038)
Mathematics
UTME / JAMB 1994
Geometry and Trigonometry
Diagram
Solution
In the diagram. Find h
Attached Figure / Diagram
A
\(\frac{12}{7}\)cm
B
\(\frac{12}{7} \sqrt{6}\)cm
C
\(\frac{7}{12}\)cm
D
\(\frac{1}{2}\)cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
A\(\bigtriangleup\) = \(\sqrt{S(S - a) (S - b)(S - c)}\) (Hero's Formula)
S = \(\frac{a + b + c}{2}\) = \(\frac{5 + 6 + 7}{2}\)
\(\frac{18}{2} = 9\)
A\(\bigtriangleup\) = \(\sqrt{9 \times 4 \times 3 \times 2}\)
\(\sqrt{216} = 6 \sqrt{6}cm^3\)
A\(\bigtriangleup\) = \(\frac{1}{2} \times 6 \times h\)
6\(\sqrt{6} = \frac{1}{2} \times 7 \times h\)
h = \(\frac{12}{7} \sqrt{6}\)cm
Source: Myschool Past Questions Archive (Record #33039)
Mathematics
WAEC / WASSCE 1994
Geometry and Trigonometry
Solution
(a) Evaluate, without using mathematical tables, \(17.57^{2} - 12.43^{2}\).
(b) Prove that angles in the same segment of a circle are equal.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(17.57^{2} - 12.43^{2}\)
Using the difference of two squares method,
= \((17.57 + 12.43)(17.57 - 12.43)\)
= \((30.00)(5.14)\)
= \(154.2\)
(b)
Given: D and C are points on the major arc of circle ADCB. To prove that < ADB = < ACB.
Construction : Join A and B to O, the centre of the circle .
Proof: With the lettering \(< AOB = 2x_{1}\) (angle at the centre is twice that subtended at the circumference)
But \(< AOB = 2x_{2}\) (the same theorem applies here)
\(\therefore 2x_{1} = 2x_{2} \implies x_{1} = x_{2}\)
\(\therefore < ADB = < ACB\)
Source: Myschool Past Questions Archive (Record #54550)
Mathematics
WAEC / WASSCE 1994
Geometry and Trigonometry
Solution
(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the tower is 65^\circ. From the top Q of the building, the angle of elevation of the point T is 25^\circ. If the building is 20m high, calculate the distance PT.
(b) Hence or otherwise, calculate the height of the tower. [Give your answers correct to 3 significant figures].
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
\(< QPB = 90^\circ - 65^\circ = 25^\circ\)
Also, \(< TQP = 90^\circ + 25^\circ = 115^\circ\)
\(\therefore < QTP = 180^\circ - (25^\circ + 115^\circ) = 40^\circ\)
Using sine rule, \(\frac{PQ}{\sin 40^\circ} = \frac{PT}{\sin 115^\circ}\)
\(PT = \frac{20 \sin 115^\circ}{\sin 40^\circ} = \frac{20 \sin 65^\circ}{\sin 40^\circ}\)
= \(28.199m \approxeq 28.2m\) (to 3 sig. fig)
(b) Height of the tower = ST
\(\sin 65 = \frac{ST}{PT} = \frac{ST}{28.199}\)
\(ST = 28.199 \times \sin 65^\circ = 25.557m\)
\(\approxeq 25.6m\) (to 3 sig. figs)
Source: Myschool Past Questions Archive (Record #54551)
Mathematics
WAEC / WASSCE 1994
Geometry and Trigonometry
Diagram
Solution
(a) Using a ruler and a pair of compasses only, construct (i) a triangle XYZ in which /YZ/ = 8cm, < XYZ = 60^\circ and < XZY = 75^\circ. Measure /XY/; (ii) the locus \(l_{1}\) of points equidistant from Y and Z ; (iii) the locus \(l_{2}\) of points equidistant from XY and YZ.
(b) Measure QY where Q is the point of intersection of \(l_{1}\) and \(l_{2}\).
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
(b) /QY/ = 4.8 cm
Source: Myschool Past Questions Archive (Record #54556)
Mathematics
WAEC / WASSCE 1994
Geometry and Trigonometry
Solution
(a) Two points X(32^\circN, 47^\circW) and Y(32^\circN, 25^\circE) are on the earth's surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour. [Radius of the earth = 6400km; \(\pi = \frac{22}{7}\)]
(b) Two observers P and Q, 15metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q are 35^\circ and 45^\circ respectively. Find the height of the kite to the nearest metre.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
Difference in longitude = 47^\circ + 25^\circ = 72^\circ
Distance /XY/ = \(\frac{\theta}{360^\circ} \times 2\pi r\)
Where \(r = R \cos 32^\circ\)
\(\therefore /XY/ = \frac{72}{360} \times 2 \times \frac{22}{7} \times 6400 \times \cos 32\)
= \(2 \times \frac{22}{7} \times 1280 \times 0.8480\)
= \(6822.765km\)
\(Speed = \frac{Distance}{Time}\)
= \(\frac{6822.766}{11} = 620.25 km/h\)
(b) \
\(
Source: Myschool Past Questions Archive (Record #54560)
Mathematics
UTME / JAMB 1993
Geometry and Trigonometry
Diagram
Solution
From the figure, calculate TH in centimeters
Attached Figure / Diagram
A
\(\frac{5}{\sqrt{3} + 1}\)
B
\(\frac{5}{\sqrt{3} - 1}\)
C
\(\frac{5}{\sqrt{3}}\)
D
\(\frac{\sqrt{3}}{5}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
TH = tan 45o, TH = QH
\(\frac{TH}{5 + QH}\) = tan 30o
TH = (b + QH) tan 30o
QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)
QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)
QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)
= \(\frac{5}{\sqrt{3} - 1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #33020)
Mathematics
WAEC / WASSCE 1993
Geometry and Trigonometry
Solution
A sector of a circle of radius 9cm subtends angle 120^\circ at the centre of the circle. Find the area of the sector to the nearest cm\(^2\) [Take \pi = 22/7]
A
75cm2
B
84cm2
C
85cm2
D
86cm2
E
95cm2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of a sector = \theta/360 x \pir\(^2\)
= 120/360 x 22/7 x 81/1
= 84.86 cm\(^2\)
\(\approxeq\) 85cm\(^2\) (to the nearest cm)
Source: Myschool Past Questions Archive (Record #23200)
Mathematics
WAEC / WASSCE 1993
Geometry and Trigonometry
Solution
The area and a diagonal of a rhombus are 60 cm\(^2\) and 12 cm respectively. Calculate the length of the other diagonal.
A
5cm
B
6cm
C
10cm
D
12cm
E
15cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of rhombus = \(\frac{pq}{2}\)
where p and q are the two diagonals of the rhombus.
\(\therefore 60 = \frac{12 \times q}{2}\)
6q = 60 \(\implies\) q = 10 cm
Source: Myschool Past Questions Archive (Record #23203)
Mathematics
WAEC / WASSCE 1993
Geometry and Trigonometry
Solution
The angle of a sector of a circle radius 10.5cm is 120^\circ. Find the perimeter of the sector [Take \pi = 22/7]
A
22cm
B
33.5cm
C
43cm
D
66cm
E
115.5cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Perimeter of a sector = \(2r + \frac{\theta}{360} \times 2\pi r\)
= \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\)
= \(21 + 22\)
= 43 cm
Source: Myschool Past Questions Archive (Record #23204)
Mathematics
WAEC / WASSCE 1993
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQR is a tangent to the circle QST at Q. If |QT| = |ST| and ∠SQR = 68^\circ, find ∠PQT.
Attached Figure / Diagram
A
34o
B
48o
C
56o
D
68o
E
73o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
< STQ = < SQR = 68^\circ (alternate segment)
\(\therefore\) < STQ = 68^\circ
< TQS = \(\frac{180^\circ - 68^\circ}{2}\)
= \(\frac{112}{2} = 56^\circ\)
\(\therefore\) < PQT = 180^\circ - (68^\circ + 56^\circ)
= 180^\circ - 124^\circ
= 56^\circ
Source: Myschool Past Questions Archive (Record #23205)
Mathematics
WAEC / WASSCE 1993
Geometry and Trigonometry
Solution
The sum of an interior angles of a regular polygon is 30 right angles. How many sides has the polygon?
A
34 sides
B
30 sides
C
26 sides
D
17 sides
E
15 sides
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Sum of interior angles in a polygon = \((2n - 4) \times 90^\circ\)
\(\therefore (2n - 4) \times 90^\circ = 30 \times 90^\circ\)
\(\implies 2n - 4 = 30 \)
\(2n = 34 \implies n = 17\)
The polygon has 17 sides.
Source: Myschool Past Questions Archive (Record #23206)
Mathematics
WAEC / WASSCE 1993
Geometry and Trigonometry
Diagram
Solution
In the diagram above, |QR| = 12cm and |QS| = 10cm. If ∠PQR = 90^\circ, ∠RSQ = 90^\circ and PSQ is a straight line, find |PS|
Attached Figure / Diagram
A
3.3cm
B
4.4cm
C
5cm
D
5.5cm
E
6.6cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{PQ}{RQ} = \frac{RQ}{SQ}\) (Similar triangles)
\(\therefore \frac{PS + 10}{12} = \frac{12}{10}\)
\(10 (PS + 10) = 12 \times 12\)
\(10 PS + 100 = 144 \implies 10 PS = 44\)
\(PS = 4.4 cm\)
Source: Myschool Past Questions Archive (Record #23207)