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Mathematics
WAEC / WASSCE 1995
Geometry and Trigonometry
The locus of a point which is equidistant from two given fixed points is the
A
perpendicular bisector of the straight line joining them
B
angle bisector of the straight lines joining the points to the origin
C
perpendiculars to the straight line joining them
D
parallel-line to the straight line joining them
E
a line making an acute angle with the line joining the two points
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 #23779)
Mathematics
WAEC / WASSCE 1995
Geometry and Trigonometry
Diagram
Solution
In the diagram above, the value of angles b + c is
Attached Figure / Diagram
A
180o
B
90o
C
45o
D
ao
E
do
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
a + c + d = 180o[sum of angles in triangle]
x = a[vertical opp. ∠s]
y + x + d = 180o[sum of ∠s in a \Delta]
y + a + d = 180o
y + a + d = a + c + d
y = c
but b + y = 180o [sum of ∠s on a straight line ]
b + c = 180o; Ans = A = 180o
Source: Myschool Past Questions Archive (Record #23780)
Mathematics
WAEC / WASSCE 1995
Geometry and Trigonometry
Solution
Which of the following angles is an exterior angle of a regular polygon?
A
95o
B
85o
C
78o
D
75o
E
72o
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
The formula for the exterior angle of a regular polygon = \(\frac{360}{n}\).
Among the options, only 72^\circ is divisible by 360^\circ to give an integer.
Source: Myschool Past Questions Archive (Record #23781)
Mathematics
WAEC / WASSCE 1995
Geometry and Trigonometry
Solution
Without using tables, find the value of \(\frac{\sin 20^\circ}{\cos 70^\circ} + \frac{\cos 25^\circ}{\sin 65^\circ}\)
A
2
B
1
C
o
D
-1
E
-2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Note: sin 20^\circ = cos 70^\circ hence
\(\frac{\sin 20}{\cos 70} = 1\)
Also, cos 65^\circ = sin 25^\circ
\(\frac{\cos 65}{\sin 25} = 1\)
\(\therefore \frac{\sin 20}{\cos 70} + \frac{\cos 65}{\sin 25}\)
= 1 + 1
= 2
Source: Myschool Past Questions Archive (Record #23783)
Mathematics
WAEC / WASSCE 1995
Geometry and Trigonometry
Diagram
Solution
In the diagram above, ∠PRQ = 90^\circ, ∠QPR = 30^\circ and /PQ/ = 10 cm. Find y.
Attached Figure / Diagram
A
6cm
B
5cm
C
4cm
D
3cm
E
2cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\sin 30^\circ = \frac{y}{10}\)
\(y = 10 \sin 30^\circ = 10 \times 0.5 = 5 cm\)
Source: Myschool Past Questions Archive (Record #23784)
Mathematics
WAEC / WASSCE 1995
Geometry and Trigonometry
Solution
The bearing of two points Q and R from a point P are 030^\circ and 120^\circ respectively, lf /PQ/ = 12 m and /PR/ = 5 m, find the distance QR.
A
13m
B
11m
C
9m
D
7m
E
5m
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x\(^2\) = 12\(^2\) + 5\(^2\)
x\(^2\) = 144 + 25
x\(^2\) = 169
x = \(\sqrt{169}\) = 13m.
Source: Myschool Past Questions Archive (Record #23785)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Solution
Express the product of 0.0014 and 0.011 in standard form
A
1.54 x 10-2
B
1.54 x 10-3
C
1.54 x 10-4
D
1.54 x 10-5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(0.0014 = 1.4 \times 10^{-3}\)
\(0.011 = 1.1 \times 10^{-2}\)
\(\therefore 0.0014 \times 0.011 = 1.4 \times 10^{-3} \times 1.1 \times 10^{-2}\)
= \(1.54 \times 10^{-5}\)
Source: Myschool Past Questions Archive (Record #30877)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
In a triangle XYZ, < YXZ = 44^\circ and < XYZ = 112^\circ. Calculate the acute angle between the internal bisectors of < XYZ and < XZY.
A
12o
B
56o
C
68o
D
78o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #30925)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Solution
Find the distance between two towns P(45^\circN, 30^\circW) and Q(15^\circS, 30^\circW) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)]
A
\(\frac{11000}{3}\)km
B
\(\frac{2200}{3}\)km
C
\(\frac{22000}{3}\)km
D
\(\frac{1100}{3}\)km
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Angular difference = 45^\circ + 15^\circ = 60^\circ.
\(D = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7000\)
= \(\frac{22000}{3} km\)
Source: Myschool Past Questions Archive (Record #30930)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Solution
P is on the locus of points equidiatant from two given points X and Y. UV is a straight line throuh Y parallel to the locus. If < PYU is 40^\circ, find < XPY.
A
100o
B
80 o
C
50 o
D
40 o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
< PYU = < YPQ = 40^\circ(Alternative) = PQ\(\parallel\)UV
< XPQ = < YPQ = 40^\circ, Since QX = QY
therefore < XPY = 80^\circ
Source: Myschool Past Questions Archive (Record #30935)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Solution
A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank.
A
60m
B
30 \(\sqrt{3}\)m
C
20 \(\sqrt{3}\)m
D
10 \(\sqrt{3}\)m
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
h = 30 tan 60o
= 30\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #30939)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Solution
The derivative of cosec x is
A
tan x cosec x
B
-cot x cosec x
C
tan x sec x
D
-cot x sec x
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\csc x = \frac{1}{\sin x}\)
Using the quotient rule,
\(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\)
= \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\)
= \(\frac{- \cos x}{\sin^{2} x}\)
= \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\)
= \(- \cot x \csc x\)
Source: Myschool Past Questions Archive (Record #30940)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Solution
For what value of x is the tangent to the curve y = x\(^2\) - 4x + 3 parallel to the x-axis?
A
3
B
2
C
1
D
6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
At the point where the tangent is parallel to the x- axis, the slope of the curve = 0.
Given: \(y = x^{2} - 4x + 3\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x - 4\)
\(2x - 4 = 0 \implies 2x = 4 \)
\(x = 2\)
When x = 2, the tangent of the curve is parallel to the x- axis.
Source: Myschool Past Questions Archive (Record #30941)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Solution
\(\begin{array}{c|c} \text{Age in years} & 13 & 14 & 15 & 16 & 17 \\ \hline \text{No. of students} & 3 & 10 & 30 & 42 & 15\end{array}\)
The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is
A
27o
B
30o
C
54o
D
108o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Number of 15- year old pupils = 30.
Total number of students: 3 + 10 + 30 + 42 + 15 = 100.
Angle = \(\frac{30}{100} \times 360^\circ\)
= \(108^\circ\)
Source: Myschool Past Questions Archive (Record #30944)
Mathematics
UTME / JAMB 1995
Geometry and Trigonometry
Diagram
Solution
Determine the value of x in the figure
Attached Figure / Diagram
A
134o
B
81o
C
53o
D
46o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Within the triangle,
180^\circ - (81^\circ + 53^\circ) = 46^\circ
In a cyclic quadrilateral, the sum of two opposite angles = 180^\circ
\(\therefore\) x = 180^\circ - 46^\circ = 134^\circ
Source: Myschool Past Questions Archive (Record #33044)