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Mathematics
WAEC / WASSCE 2017
Geometry and Trigonometry
Solution
(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, < TPS = 34^\circ and < TSQ = 65^\circ.
(i) Illustrate the information in a diagram; (ii) find the value of : (a) < RTS ; (b) < SRP.
(b)
In the diagram, /VZ/ = /YZ/, < YXZ = 20^\circ and < ZVY = 52^\circ. Calculate the size of < WYZ.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(ii) (a) < RTS = 65^\circ - 34^\circ (exterior angles of triangle PST)
= 31^\circ
(ii) (b) < SRP = 65^\circ (angles in alternate segment are equal)
\(\therefore < SRP = 180^\circ - 65^\circ = 115^\circ\).
(b)
If < XYZ = 128^\circ
then < XZY = 128^\circ + 20^\circ = 148^\circ
< XZV = 180^\circ - 148^\circ = 32^\circ
< YVZ = < VYZ = 52^\circ (Isoceles triangle)
< WYZ = 52^\circ - 32^\circ = 20^\circ
Source: Myschool Past Questions Archive (Record #52358)
Mathematics
WAEC / WASSCE 2017
Geometry and Trigonometry
Solution
(a) Given that \(\sin x = \frac{5}{13}, 0^\circ < x < 90^\circ\), find \(\frac{\cos x - 2\sin x}{2\tan x}\).
(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6metres above the groung. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the :
(i) length of ladder LA (ii) angle which LA makes with the ground.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\sin x = \frac{5}{13}\)
Using SOHCAHTOA, then Opp = 5 and Hyp = 13
\(13^{2} = 5^{2} + Adj^{2}\)
\(169 = 25 + Adj^{2}\)
\(Adj = \sqrt{169 - 25} = \sqrt{144} = 12\)
\(\cos x = \frac{12}{13}\)
\(\tan x = \frac{5}{12}\)
\(\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - 2(\frac{5}{13})}{2(\frac{5}{12})}\)
= \(\frac{\frac{2}{13}}{\frac{5}{6}}\)
= \(\frac{12}{65}\)
(b)
Taking \(\Delta LPB\),
Using Pythagoras theorem, \(/BP/^{2} = /BL/^{2} - /LP/^{2}\)
\(i.e. /BP/ = \sqrt{12^{2} - 9.6^{2}}\)
= \(\sqrt{144 - 92.16} = \sqrt{51.84} = 7.2m\)
\(/PA/ = 10m - 7.2m = 2.8m\)
\(/LA/^{2} = x^{2}\)
\(x^{2} = /LP/^{2} + /PA/^{2}\)
\(x^{2} = 9.6^{2} + 2.8^{2}\)
= \(92.16 + 7.84\)
\(x^{2} = 100 \implies x = 10m\)
(ii) \(\tan \theta_{y} = \frac{9.6}{2.8} = 3.4286\)
\(\theta_{y} = \tan^{-1} (3.4286) \approxeq 74^\circ\)
Source: Myschool Past Questions Archive (Record #52375)
Mathematics
WAEC / WASSCE 2017
Geometry and Trigonometry
Solution
(a) It takes 8 students two- thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one- third of an hour at the same rate?
(b) A chord, 20 cm long, is 12 cm from the centre of the circle. Calculate, correct to one decimal place, the :
(i) angle subtended by the chord at the centre of the circle;
(ii) perimeter of the minor segment cut off by the chord. [Take \(\pi = 3.142\)].
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(Time = \frac{2}{3} hour = \frac{2}{3} \times 60 = 40 mins\)
4 students will take \(\frac{8 \times 40}{4} = 80 minutes\)
\(Time = \frac{1}{3} hour = \frac{1}{3} \times 60 = 20 minutes\)
It will take 20 minutes to fill \(\frac{12 \times 20}{80} = 3 tanks\)
(b)
\(\tan (\frac{\theta}{2}) = \frac{10}{12}\)
\(\frac{\theta}{2} = \tan^{-1} (0.8337) = 39.8^\circ\)
\(\theta = 39.8^\circ \times 2 = 79.6^\circ\)
(ii) Perimeter of minor segment = Length of chord + length of arc
But \(r^{2} = 12^{2} + 10^{2}\)
= \(144 + 100 = 244\)
\(r = \sqrt{244} = 15.62 cm\)
Length of arc = \(\frac{79.6}{360} \times 2 \times 3.142 \times 15.62 = 21.71cm\)
Perimeter of minor segment = 20 + 21.71 = 41.71cm
Source: Myschool Past Questions Archive (Record #52385)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
The sum of the interior angles of a polygon is a given as 1080o. Find the number of the sides of the polygon.
A
5
B
7
C
6
D
8
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{(n - 2) 180}{180}\) = \(\frac{1080}{180}\)
(n - 2) = 6
n = 6 + 2 = 8
Source: Myschool Past Questions Archive (Record #44750)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Diagram
Solution
In the diagram above, l1 is parallel to l2, Find the value of < PMT
Attached Figure / Diagram
A
82o
B
36o
C
72o
D
118o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
o - 118o = 62o
o ( Alternative angle)
y + 2y + 10o + 62o = 180o (Angles on a straight line)
3y = 180 - 72
\(\frac{3y}{3}\) = \(\frac{108}{3}\)
y = 36o
o
Source: Myschool Past Questions Archive (Record #44751)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Diagram
Solution
From the diagram above, Find the value of < ROP
Attached Figure / Diagram
A
110o
B
70o
C
95o
D
85o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
o (Exterior angle of cyclic quadrilateral)
Source: Myschool Past Questions Archive (Record #44755)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
Evaluate \(\int\)(cos4x + sin3x)dx
A
sin4x - cos3x + k
B
sin4x + cos3x + k
C
\(\frac{1}{4}\)sin4x - \(\frac{1}{3}\)cos3x + k
D
\(\frac{1}{4}\)sin4x + \(\frac{1}{3}\)cos3x + k
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\int (\cos 4x + \sin 3x) \mathrm d x\)
= \(\frac{1}{4} \sin 4x - \frac{1}{3} \cos 3x + k\)
Source: Myschool Past Questions Archive (Record #44769)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Diagram
Solution
The pie chart above shows the distribution of subjects offered by students in SSS III level. If 80 students enrolled in the class. What is the size of the angle of the sector in economics?
Attached Figure / Diagram
A
24o
B
39o
C
32o
D
36o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
xo + 36o + 54o + 28o + 90o + 120o = 360o
x + 328o = 360o
x = 360o - 328o
x = 32o
Source: Myschool Past Questions Archive (Record #44771)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
Calculate the perimeter of a sector of a circle of raduis 12cm and angle 60o.
A
(12 + 4\(\pi\))cm
B
(24 + 4\(\pi\))cm
C
(12 + 6\(\pi\))cm
D
(24 + 6\(\pi\))cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Perimeter = 2r + \(\frac{\theta}{360}\) x 2\(\pi\)r
= 2(12) + \(\frac{60}{360}\) x 2 x \(\pi\) x 12
= (24 + 4\(\pi\))cm
Source: Myschool Past Questions Archive (Record #44775)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
Simplify \(\frac{0.026 \times 0.36}{0.69}\). Leave your answer in standard form
A
1.36 x 10-4
B
1.36 x 10-3
C
1.36 x 10-1
D
1.36 x 10-2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{0.026 \times 0.36}{0.69}\) = \(\frac{26 \times 10^-3 \times 36 \times 10^-2}{69 \times 10^-2}\)
= 0.01356 = 1.36 x 10-2
Source: Myschool Past Questions Archive (Record #44827)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
Calculate the perimeter of a sector of a circle of raduis 9cm and angle 36o.
A
18cm
B
(18 + \(\frac{9\pi}{5}\))cm
C
(9 + \(\frac{9\pi}{5}\))cm
D
\(\frac{9\pi}{5}\)cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Perimeter = 2r + \(\frac{\theta}{360}\) x 2\(\pi\)r
2(9) + \(\frac{36}{360}\) x 2 x \(\pi\) x 9 = 18 + \(\frac{9\pi}{5}\)
Source: Myschool Past Questions Archive (Record #44829)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
An arc subtends an angle of 30º at the centre of a circle radius 12cm. Calculate the length of the arc.
A
6\(\pi\)cm
B
2\(\pi\)cm
C
3\(\pi\)cm
D
9\(\pi\)cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Length or arc = \(\frac{\theta}{360}\) x 2\(\pi\)r
= \(\frac{30}{360}\) x 2 x \(\frac{22}{7}\) x 12
= \(\frac{1}{12}\) x 2 x \(\pi\) x 12 = 2\(\pi\)cm
Source: Myschool Past Questions Archive (Record #44834)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
Evaluate \(\int\limits_0^\frac{\pi}{2}\) sin xdx
A
2
B
-1
C
1
D
-2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\int\limits_0^\frac{\pi}{2}\) sin xdx = -cosx|\(\frac{\pi}{2}\)
= -(cos\(\frac{\pi}{2}\) - cos0) = -(0-1) = 1
Source: Myschool Past Questions Archive (Record #44835)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
An arc of the length 16\(\pi\)cm subtends an angle of 80o at the centre of the circle. Find the radius of the circle.
A
24cm
B
28cm
C
36cm
D
32cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
L = \(\frac{\theta}{360}\) x 2\(\pi\)r
16\(\pi\) = \(\frac{80}{360}\) x 2\(\pi\)r
\(\frac{16\pi \times 360}{80 \times 2\pi}\) = r
r = 36cm
Source: Myschool Past Questions Archive (Record #44836)
Mathematics
UTME / JAMB 2016
Geometry and Trigonometry
Solution
Evaluate \(\int\)(sinx - 5x2)dx
A
-cosx - 10x + k
B
cosx - \(\frac{5x^3}{3}\) + k
C
-cosx - \(\frac{5x^3}{3}\) + k
D
cosx - 10x + k
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\int\)(sin x - 5x^2) = -cosx - \(\frac{5x^3}{3}\) + k
Source: Myschool Past Questions Archive (Record #44861)