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Mathematics
UTME / JAMB 2008
Geometry and Trigonometry
Solution
The locus of a point equidistant from two points p(6,2) and R(4,2) is a perpendicular bisector of PR passing through?
A
(2,5)
B
(5,2)
C
(1,0)
D
(0,1)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
X = \(\frac{6+4}{2}\) = \(\frac{10}{2}\) = 5
Y = \(\frac{2+2}{2}\) = \(\frac{4}{2}\) =2
∴ X, Y = (5,2)
Source: Myschool Past Questions Archive (Record #9284)
Mathematics
UTME / JAMB 2008
Geometry and Trigonometry
Solution
If sin\theta = 3/5. Find Tan\theta
A
3/4
B
3/5
C
2/5
D
1/4
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
sin\theta = 3/5
x2 = 52 - 32
Source: Myschool Past Questions Archive (Record #9286)
Mathematics
UTME / JAMB 2008
Geometry and Trigonometry
Solution
Differentiate sin x - x cos x
A
x cos x
B
x sin x
C
-x cos x
D
-x sin x
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
sin x - x cos x
dy/dx = cos x - [1.cos x + x -sin x]
= co x - [cos x - x sin x]
= cos x - cos x + x sin x
= x sin x
Source: Myschool Past Questions Archive (Record #9289)
Mathematics
UTME / JAMB 2008
Geometry and Trigonometry
Solution
Evaluate \(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx\)
A
zero
B
1
C
2
D
3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx = [sinx]^{\frac{\pi}{2}} _{\frac{-\pi}{2}}\)
=sin\(\frac{\pi}{2} - sin\frac{-\pi}{2}\)
= sin90 – sin(-90)
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2
Source: Myschool Past Questions Archive (Record #9292)
Mathematics
UTME / JAMB 2008
Geometry and Trigonometry
Solution
If x > 0, find the range of number x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x and 3x+1
A
3x+3
B
3x+1
C
2x+5
D
2x+1
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x, 3x+1
Range = 3x + 2 - (x-3)
= 2x + 5
(Assuming x = 1, since x > 0)
Source: Myschool Past Questions Archive (Record #9297)
Mathematics
UTME / JAMB 2008
Geometry and Trigonometry
Solution
In how many ways can the letters of the word ACCEPTANCE be arranged?
A
\(\frac{10!}{(2!2!3!)}\)
B
\(\frac{10!}{( 2!3!)}\)
C
\(\frac{10!}{(2!2!)}\)
D
10!
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
ACCEPTANCE = 10 Letters
A = 2 letters
C = 3 letters
E = 2 letters
It can be arranged in \(\frac{10!}{(2!3!2!)}\) ways
Source: Myschool Past Questions Archive (Record #9299)
Mathematics
UTME / JAMB 2008
Geometry and Trigonometry
Solution
A student sitting on a tower 68 metres high observes his principal's car at the angle of depression of 20o. How far is the car from the bottom of the tower to the nearest metre?
A
184m
B
185m
C
186m
D
187m
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Tan 20º = \(\frac{68m}{x}\)
x tan 20º = 68
x = \(\frac{68}{tan 20}\) = \(\frac{68}{0.364}\)
x = 186.8
= 187m
Source: Myschool Past Questions Archive (Record #29982)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Solution
PQR is a sector of a circle centre O, radius 4cm. If PQR = 30o, find, correct to 3 significant figures, the area of sector PQR. [Take \(\pi = \frac{22}{7}\)]
A
4.19cm2
B
8.38cm2
C
10.5cm2
D
20.9cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Given q2 = 25 - r2.....(1)
from pythagora's theorem
P2 = q2 + r2.......(2)
put (1) into (2)
p2 = 25 - p2 + x2
p2 = 25
p = \(\sqrt{25}\)
= 5
Source: Myschool Past Questions Archive (Record #36358)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Solution
The angles of a quadrilateral are (x + 10)o, 2yo, 90o and (100 - y)o, Find y in terms of x
A
y = 160 + x
B
y = 100 + x
C
y = 160 - x
D
y = x - 100
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Sum of the angles in a quadrilateral = 360o
(x + 10o) + 2y + 90o + 100 - y = 360o
x + y + 200 = 360o
y = 360 - 200 - x
y = 160 - x
Source: Myschool Past Questions Archive (Record #36361)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Solution
If xo is obtuse, which of the following is true?
A
x . 90
B
180 < x < 270
C
x < 90
D
90 < x < 180
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
obtuse angle \(\to\) 90o o
Source: Myschool Past Questions Archive (Record #36362)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Solution
If cos (x + 25)o = sin 45o, find the value of x
A
20
B
30
C
45
D
60
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
cos(x + 25o) = sin 45o
using cos \(\theta\) = sin(90 - \(\theta\))
cos(x + 25o) = cos(90 - 45)
cos(x + 25o) = cos 45
x + 25 = 45
x = 45 - 25
x = 20o
Source: Myschool Past Questions Archive (Record #36364)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Solution
A train travels 60km in M minutes. If its average speed is 400km per hour, find the value of M
A
15
B
12
C
10
D
9
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Average speed = \(\frac{Distance}{Time}\)
\(\frac{400km}{hr} = \frac{60km}{Time}\)
Time = \(\frac{60km}{400 km/hr}\)
= \(\frac{60hr}{400}\)
M = \(\frac{60hr}{400} \times \frac{60min}{1hr}\)
= 9 minutes
Source: Myschool Past Questions Archive (Record #36381)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Solution
An arc of a circle, radius 14cm, is 18.33cm long. Calculate to the nearest degree, the angle which the arc subtends at the centre of the circle. [T = \(\frac{22}{7}\)]
A
11o
B
20o
C
22o
D
75o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Length of an arc = \(\frac{\theta}{360} \times 2\pi r\)
18.33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 14\)
\(\theta = \frac{18.33 \times 360 \times 17}{2 \times 22 \times 14}\)
= 75o (approx.)
Source: Myschool Past Questions Archive (Record #36382)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Solution
XY is a chord of circle centre O and radius 7cm. The chord XY which is 8cm long subtends an angle of 120o at the centre of the circle. Calculate the perimeter of the minor segment. [Take \(\pi = \frac{22}{7}\)]
A
14.67cm
B
22.67cm
C
29.33cm
D
37.33cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
perimeter of minor segment = Length of arc xy + chord xy
where lxy = \(\frac{120}{360} \times 2x \times \frac{22}{7} \times 7\)
= 14.67cm
perimeter of minor segment = 14.67 + 8 = 22.67cm
Source: Myschool Past Questions Archive (Record #36386)
Mathematics
WAEC / WASSCE 2008
Geometry and Trigonometry
Diagram
Solution
In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS
Attached Figure / Diagram
A
150o
B
120o
C
90o
D
60o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Since |PR| = |RS| = |SP|
\(\bigtriangleup\) PRS is equilateral and so o
But o (Opposite interior angles of a cyclic quadrilateral)
o
o
But in \(\bigtriangleup\) PQR, PQ = PR, hence o (Angles in a triangle)
2 o
From the diagram, o
Source: Myschool Past Questions Archive (Record #36406)