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Mathematics
UTME / JAMB 1980
Geometry and Trigonometry
Solution
If Sin \(\theta\) = \(\frac{\sqrt{3}}{2}\) and \(\theta\) is less than 90º, calculate \(\frac{tan(90 - \theta)}{cos^2\theta}\)
A
\(\frac{1}{2}\)
B
\(\frac{4}{\sqrt{3}}\)
C
\(\sqrt{3}\)
D
4\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
If Sin \(\theta\) = \(\frac{\sqrt{3}}{2}\), then \(\theta\) = 60º
∴ \(\frac{tan(90 - \theta)}{cos^2\theta}\) = \(\frac{tan(90 - 60)}{cos^2 60}\) = \(\frac{tan30}{cos^2 60}\)
= \(\frac{1}{\sqrt3} \times (\frac{2}{1})^2 = \frac{4}{\sqrt{3}}\)
Source: Myschool Past Questions Archive (Record #32751)
Mathematics
UTME / JAMB 1980
Geometry and Trigonometry
Diagram
Solution
Find the area of the shades segments in the figure
Attached Figure / Diagram
A
\(\sqrt{3}
B
4 \(\pi - \sqrt{3}\)
C
-\(\frac{2}{3} \pi\)
D
\(\frac{2\pi}{3}\) -3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of section = \(\frac{60^o}{360^o}\) x 11r2
= \(\frac{60}{360} \times \pi \times 2^2\)
= \(\frac{1}{6}\) x 4
= \(\frac{4\pi}{6}\)
= \(\frac{2\pi}{3}\)
Area of triangle = \(\frac{1}{2x}\)
= 2 x 28.......60
Segment Area = Area of section - Area of triangle
= \(\frac{2\pi}{3}\) -3
Source: Myschool Past Questions Archive (Record #32752)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Diagram
Solution
In the figure, PQ and QR are chords of the circle PQR. QRS is a straight line and PR is equal to RS, < PSR is 20o. What is the size of
Attached Figure / Diagram
A
70o
B
90o
C
80o
D
40o
E
60o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Isosceles Triangle PSR:
RSP ≡ RPS → 20º
That is PRQ = 40º
POQ = 2 * 40 = 80º (Angle subtended by chord PQ at centre is twice angle subtended at circumference).
The size of POQ = 80º
Source: WAEC / JAMB / NECO Classroom Archive (Record #32736)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Diagram
Solution
In the parallelogram PQRS, PE is perpendicular to QR. Find the area of the parallelogram.
Attached Figure / Diagram
A
60cm2
B
65cm2
C
72cm2
D
132cm2
E
156cm2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
By Pythagoras, PE\(^2\) = 12\(^2\) - 5\(^2\)
= 144 - 25 = 119
h = PE\(^2\) = \sqrt119 = 10.9 ≈ 11cm,
Area of 11gm = b x h
QR = b =(5 + 7)cm = 12cm
area = 12 x 11
= 132cm\(^2\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #32739)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
An arc of circle of radius 2cm subtends an angle of 60º at the centre. Find the area of the sector
A
\(\frac{2 \pi}{3}\)cm2
B
\(\frac{\pi}{2}\)cm2
C
\(\frac{\pi}{3}\)cm2
D
\(\pi\)cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of a sector \(\frac{\theta}{360}\) x \(\pi\)r\(^2\)
= \(\frac{60^o}{360^o}\) x \(\pi\)2\(^2\)
= \(\frac{2 \pi}{3}\)cm2
Source: WAEC / JAMB / NECO Classroom Archive (Record #32676)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
If the value of \(\pi\) is taken to be \(\frac{22}{7}\), the area of a semi-circle of diameter 42m is
A
5544m2
B
1386m2
C
132m2
D
264m2
E
693m2
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
\(\pi\) = \(\frac{22}{7}\), Diameter D = 42m
Area of a circle = \(\frac{\pi D^2}{4}\) when a diameter is given
area of a semicircle \(\frac{\pi D^2}{4}\) x \(\frac{1}{2}\)
= \(\frac{22}{7}\) x \(\frac{5541.7694}{4}\) x \(\frac{1}{2}\)
= \(\frac{11}{7} \times 441\)
= 693m\(^{2}\)
Source: Myschool Past Questions Archive (Record #32646)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
In \(\bigtriangleup\)PQR, PQ = 10cm, QR = 8cm and RP = 6cm, the perpendicular RS is drawn from R to PQ. Find the length of RS
A
4cm
B
32cm
C
\(\frac{30}{7}\)
D
\(\frac{40}{7}\)
E
4.8cm
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Cos Q = \(\frac{r^2 + p^2 - q^2}{2rp}\)
= \(\frac{10^2 + 8^2 - 6^2}{2(10)(8)}\)
= \(\frac{164 - 36}{160}\)
= \(\frac{128}{160}\)
= 0.8
Q = Cos-1 o.8
= 37o x rep. from rto = 1
x = 8 sin 37o
x = 4.8cm
Source: Myschool Past Questions Archive (Record #32648)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
A ladder resting on a vertical wall makes an angles whose tangent is 2.4 with the ground. If the distance between the foot of the ladder and the wall is 50cm, What is the length of the ladder?
A
1m
B
1.1m
C
1.2m
D
0.9m
E
1.3m
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
If tan\(\theta\) = 2, 4
\(\theta\) = tan-1(2.4)
= 67o 25
By sin formula \(\frac{x}{sin 90^o}\) = \(\frac{50}{sin 22^o 37}\)
= \(\frac{50}{sin(9o - \theta)}\)
sin90o = 1
x = 50 cosec 22o 37
= 50 x 2.604x
= 130cm
but 100cm = 1m
130cm = 1.3m
Source: Myschool Past Questions Archive (Record #32661)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
One of the following statements is wrong. Which is it?
A
If a triangle is equiangular then it is also equilateral
B
If a triangle is equilateral then it is also then it is also equiangular
C
If two triangles are similar then they are also congurent
D
The sum of the interior angles of any triangle is 180 degrees
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Two similar triangles are not also congruent
Source: Myschool Past Questions Archive (Record #32663)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
PQRS is a cyclic quadrilateral with PQ as diameter of the circle. If o find < QRS
A
75o
B
37\(\frac{1}{2}\)o
C
127\(\frac{1}{2}\)o
D
105o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
PSO = 90o(angle in a semicircle)
SPO = 180o - (90o + 15o) = 75o
o - 75o
= 105o(opposite angles in cyclic quadrilateral are supplementary)
Source: Myschool Past Questions Archive (Record #32664)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
On each market day Mrs. Bassey walks to the market from her home at a steady speed. This journey normally takes her 2 hours to complete. She finds, however, that by increasing her usual speed by 1 km/hr she can save 20 minutes. Find her usual speed in km/hr
A
1\(\frac{2}{3}\)
B
2
C
5
D
6
E
10
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let x rep. the usual speed = \(\frac{Distance}{Time}\)
x = d/2
d = 2x ............. 1
x + 1 = \(\frac{d}{2 - 1/3}\)
x + 1 = \(\frac{3d}{5}\)
5x + 5 = 3d
d = \(\frac{5x + 5}{3}\) ................... 2
2x = \(\frac{5x + 5}{3}\)
6x = 5x + 5
x = 5km/hr
Source: Myschool Past Questions Archive (Record #32667)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
What is the greatest straight line distance between two vertices (corners) of a cube whose sides are 2239cm long?
A
\(\sqrt{2239cm}\)
B
\(\sqrt{2}\) x 2239cm
C
\(\frac{\sqrt{3}}{2}\) 2239cm
D
\(\sqrt{3}\) x 2239cm
E
4478cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x = \(\sqrt{-2239^2 + 2239^2}\)
= -\(\sqrt{10026242}\)
= 3166.42
y = -\(\sqrt{10026242 + 5013121}\)
= -\(\sqrt{15039363}\)
= 3878
= \(\sqrt{3}\) x 2239
Source: Myschool Past Questions Archive (Record #32677)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Solution
A man is standing in the corridor of a 10-storey building and looking down at a tall tree in front of the building. He sees the top of the tree at angle of depression of 30o. If the tree is 200m tall and the man's eyes are 300m above the ground, calculate the angle of depression of the foot tree as seen by the man
A
30o
B
60o
C
45o
D
25o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let x rep. the angle of depression of the foot of the tree.
tan 30o = \(\frac{y}{100}\)
y = 100 tan 30o
= 57.8
By Pythagoras, AC2 = 3002 + 582
= 900 + 3354
tan x = \(\frac{opp}{adj}\)
= \(\frac{58}{300}\)
= 0.19
tan x = 0.19
x = tan 0.19
= 60o
Source: Myschool Past Questions Archive (Record #32684)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Diagram
Solution
In the figure, \(\bigtriangleup\) ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and
Attached Figure / Diagram
A
3\(\sqrt{2}\)
B
2\(\sqrt{3}\)
C
\(\frac{\sqrt{3}}{2}\)
D
\(\frac{2}{\sqrt{3}}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
BC = 6 : DC = \(\frac{6}{2}\) = 3cm
By construction < EDE = 180o(90o + 60o) = 180o - 150o
= 30o(angle on a strt. line)
From rt < triangle ADC, AD2 = 52 - 32
= 25 - 9 = 6
AD = 4
From < AEC, let AS = x
\(\frac{x}{sin 60^o}\) - \(\frac{4}{sin 90^o}\)
sin 90o = 1
sin 60o = \(\frac{\sqrt{3}}{2}\)
x = 4sin 60o
x = 3 x \(\frac{\sqrt{3}}{2}\)
= 2\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #32733)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Diagram
Solution
In this figure, PQRS is a parallelogram, PS = PT and < PST = 55\(^o\). The size of
Attached Figure / Diagram
A
125o
B
120o
C
115o
D
110o
E
10o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Both pairs of opp. angles are equal
< STP = 55\(^o\) - isosceles angle
< TSR = 55\(^o\) - alternate angle to < STP
Hence, < PSR = 55\(^o\) + 55\(^o\) = 110\(^o\)
\(\therefore\) < PQR = 110\(^o\)
Source: Myschool Past Questions Archive (Record #32734)