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Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Diagram
In the diagram, CDE is a triangle, ABEF is cyclic quadrilateral, AB||CD, ∠BAF = 65o and ∠AFE = 85o calculate ∠CDE
Attached Figure / Diagram
A
20o
B
30o
C
50o
D
65o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #26422)
Mathematics
UTME / JAMB 1999
Geometry and Trigonometry
Diagram
Solution
In the diagram, EFGH is a cyclic quadrilateral in which EH || FG, EG, and FH are chords. If < FHG = 42º and < EFH = 34º, calculate < HEG.
Attached Figure / Diagram
A
34o
B
42o
C
52o
D
76o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
< EFH = < EGH = 34º (angles n the same segment)
< GHF = < GEF = 42º(angles in the same segment)
< FOG = 42 + 34 = 76º (exterior angle)
< FOG = < EOH = 76º (vertically opposite angle)
< EDO = 90º, < DOE = \(\frac{76}{2}\) = 38º
< HEG = 90º - 38º = 52º
Source: Myschool Past Questions Archive (Record #33085)
Mathematics
UTME / JAMB 1999
Geometry and Trigonometry
Diagram
Solution
In the figure, PQRS is a circle with ST||RQ. Find the value of x if PT = pS
Attached Figure / Diagram
A
70o
B
55o
C
40o
D
35o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
o (sum of o - 110o = 70o
in o
2x + 70 = 180o
2x = 180o - 70o = 110o
x = \(\frac{110^o}{2}\) = 55o
Source: Myschool Past Questions Archive (Record #33086)
Mathematics
UTME / JAMB 1999
Geometry and Trigonometry
Diagram
Solution
find the length XZ in the triangle
Attached Figure / Diagram
A
\(\sqrt{7m}\)
B
\(\sqrt{6m}\)
C
\(\sqrt{5m}\)
D
\(\sqrt{3m}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
xz2 = xy2 + yx2 - 2(xy) (yz) cos 120o
= 22 + 12 - 2(2) (1) cos 1202
= 4 + 1 - 4x - cos 60 = 5 - 4x - \(\frac{1}{2}\)
5 + 2 = 7
xz = \(\sqrt{7}\)m
Source: Myschool Past Questions Archive (Record #33089)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
ABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37^\circ and |BC| = 5 cm. Calculate :
(a) |AC|, correct to 3 significant figures ;
(b) < PBA.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Let |PC| = x cm; Hence, |AC| = 2x cm
\(\tan 37^\circ = \frac{x}{5}\)
\(x = 5 \tan 37 \)
\(x = 3.768 cm\)
\(\therefore |AC| = 2 \times 3.768\)
= \(7.536 cm\)
\(\approxeq 7.54 cm\) (3 sig. figs)
(b) From \(\Delta ABC\),
\(\tan < ABC = \frac{7.536}{5} = 1.5072\)
\(< ABC = \tan^{-1} (1.5072) = 56.436^\circ\)
\(\therefore < PBA = < ABC - < PBC\)
= \(56.436^\circ - 37^\circ\)
= \(19.436^\circ \approxeq 19.44^\circ\)
Source: Myschool Past Questions Archive (Record #54417)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :
(a) |AB| ;
(b) the area of the triangle BCD ;
(c) |CD| ;
(d) perimeter of the trapezium.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(|AB| = \sqrt{|BD|^{2} - |AD|^{2}}\)
= \(\sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225}\)
= \(\sqrt{64} = 8 cm\)
(b) Area of triangle BCD = \(\frac{1}{2} \times 9 \times 8 \)
= \(36 cm^{2}\)
(c) \(|EC| = 6 cm\)
\(|CD| = \sqrt{8^{2} + 6^{2}} = \sqrt{100}\)
= 10 cm.
(d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm
= 42 cm.
Source: Myschool Past Questions Archive (Record #54418)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\(^{3}\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \(\pi = \frac{22}{7}\)]
(b) In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) Volume of outer pipe = \(\pi r^{2} h\)
= \(\frac{22}{7} \times (5.0)^{2} \times 28 \times 100\)
Volume of inner pipe = \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)
Volume of metal used = Volume of outer pipe - Volume of inner pipe
= \(22 \times 400 \times (5.0^{2} - 3.5^{2})\)
= \(22 \times 400 \times (5.0 - 3.5) \times (5.0 + 3.5)\)
= \(112,200 cm^{3}\)
(ii) Vol of water the pipe can hold = vol of inner pipe
= \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)
= \(107,800 cm^{3}\)
= 107.8 litres.
(b) < PMN = < LNM (alternate angles)
< PMN = < LMN (angles in alternate segment)
\(\therefore\) < LNM = < NLM, then \(\Delta\) LMN is isosceles.
Source: Myschool Past Questions Archive (Record #54422)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70^\circ to the horizontal.
(a) How high up the wall, correct to the nearest metre, does the ladder reach?
(b) If the foot (F) of the ladder is pulled from the wall to F\(^{1}\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\(^{1}\).
(ii) Calculate, correct to the nearest degree, \(QF^{1}T^{1}\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(TQ = 10 \sin 70^\circ = 10 \times 0.9397 = 9.397m\)
\(\approxeq 9m\)
(b)(i) \(QF = 10 \cos 70^\circ = 10 \times 0.3420 = 3.42m\)
\(QF^{1} = 3.42m + 1m = 4.42m\)
\(QT^{1} = \sqrt{10^{2} - 4.42^{2}} = \sqrt{100 - 19.5364}\)
= \(\sqrt{80.4636} = 8.97m\)
\(TT^{1} = 9.397 - 8.97 = 0.43m\)
(ii) \(\sin < QF^{1}T^{1} = \frac{8,97}{10} = 0.897\)
\(< QF^{1} T^{1} = 63.768^\circ \approxeq 64^\circ\)
Source: Myschool Past Questions Archive (Record #54424)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.
(a) Find the cost when there are 44 pupils.
(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Let P represent the cost ; n, the number of pupils and k and c, constant terms.
\(\therefore P = k + cn\)
\(15705 = k + 50c ... (1)\)
\(13305 = k + 40c .... (2)\)
(1) - (2) :
\(2400 = 10c \implies c = \frac{2400}{10} = 240\)
Put c = 240 in (1), so that
\(15705 = k + (50)(240)\)
\(15705 = k + 12000 \implies k = 3705\)
\(P = 3705 + 240n\) ------ formula connecting price and number of pupils.
(a) When number of pupils, n = 44
\(P = 3705 + 240(44) = 3705 + 10560 = $14265\)
(b) For school not to be at loss:
\(360 \geq k + cn\)
\(360n \geq 3705 + 240n\)
\(360n - 240n \geq 3705 \implies 120n \geq 3705\)
\(n \geq \frac{3705}{120}\)
\(n \geq 30.875 \)
\(n \geq 31\)
The pupils should be 31 at least for the school not to run at loss.
Source: Myschool Past Questions Archive (Record #54425)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
On a graph sheet, using a scale of 2cm to 2 units on both axes,
(a) Draw the straight line joining points P(-5, 3) and Q(2, 3);
(b) construct the locus L of points equidistant from P and Q;
(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;
(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\(^{2}\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Coordinate of R and S : R(-1.6, 6.5) and S(-1.6, 0.6).
Area of rhombus PRQS : \(\frac{(5 + 2) \times (6.5 + 0.6)}{2} = \frac{49.7}{2}\)
= \(24.85 cm^{2}\)
Source: Myschool Past Questions Archive (Record #54426)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
(a) A shop owner marked a shirt at a price to enable him to make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer at N864.00. What was the original cost to the shop owner?
(b) A rectangular lawn of length (x + 5) metres is (x - 2) metres wide. If the diagonal is (x + 6) metres, find ;
(i) the value of x ; (ii) the area of lawn.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Price with 20% gain = 100% = x
Selling price = 100% - 10% = 90%
i.e. 90% of x = 864
\(\therefore x = 864 \times \frac{100}{90} = N960\)
Let cost price = y = 100%
x = 20% of y + y = 120% of y.
\(y = \frac{100}{120} x = \frac{100}{120} \times N960 = N800\)
(b)
(i) \((x + 6)^{2} = (x + 5)^{2} + (x - 2)^{2}\)
\(x^{2} + 12x + 36 = x^{2} + 10x + 25 + x^{2} - 4x + 4\)
\(x^{2} + 12x + 36 = 2x^{2} + 6x + 29\)
\(2x^{2} - x^{2} + 6x - 12x + 29 - 36 = 0\)
\(x^{2} - 6x - 7 = 0\)
\(x^{2} - 7x + x - 7 = 0 \implies x(x - 7) + 1(x - 7) = 0\)
\((x - 7)(x + 1) = 0 \implies \text{x = 7 or -1}\)
Since measurements cannot be negative, then x = 7 is the suitable answer.
(ii) Length of the lawn = (x + 5) metres = (7 + 5) = 12 metres.
Width of the lawn = (x - 2) metres = (7 - 2) = 5 metres
\(\therefore \text{The area of the lawn} = 12 \times 5 = 60 m^{2}\)
Source: Myschool Past Questions Archive (Record #54427)
Mathematics
WAEC / WASSCE 1999
Geometry and Trigonometry
Solution
In the diagram above, RST is a tangent to circle VSU center O ∠SVU = 50^\circ and UV is a diameter. Calculate ∠RSV.
A
90°
B
50°
C
45°
D
40°
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
< VSU = 90^\circ (angle in a semi-circle)
\(\therefore\) < SUV = 180^\circ - 90^\circ - 50^\circ
= 40^\circ
< RSV = 40^\circ (angle which chord makes tangent = angle in alternate segment)
Source: Myschool Past Questions Archive (Record #54932)
Mathematics
WAEC / WASSCE 1998
Geometry and Trigonometry
Solution
Cos x is negative and sin x is negative.Which of the following is true of x?
A
0o o
B
90o o
C
180o o
D
270o o
E
-90o o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the 3rd quadrant, only the tan of angles are positive.
The 3rd quadrant = 180^\circ < x < 270^\circ
Source: Myschool Past Questions Archive (Record #22096)
Mathematics
WAEC / WASSCE 1998
Geometry and Trigonometry
Solution
When an aeroplane is 800m above the ground, its angle of elevation from a point P on the ground is 30o. How far is the plane from P by line of sight?
A
400m
B
800m
C
1500m
D
1600m
E
1700m
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
From the diagram, \(\sin 30 = \frac{800}{x}\)
\(x = \frac{800}{\sin 30} \)
= \(\frac{800}{0.5} \)
= 1600 m
Source: Myschool Past Questions Archive (Record #22100)
Mathematics
UTME / JAMB 1998
Geometry and Trigonometry
Diagram
Solution
In diagram, PQ || ST and < PQR = 120º, < RST = 130º, find the angle marked x
Attached Figure / Diagram
A
50
B
65
C
70
D
80
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Produce line PQ to meet line RS at U
< RQU = 180º - 120º = 60º (
In angle QRU: < QPT + RQU = < RUV ( 2 interior angles = one exterior angle)
x + 60º = 130º and x = 70º
Source: WAEC / JAMB / NECO Classroom Archive (Record #33077)