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Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
if tanθ = \(frac{3}{4}\), 180° < θ < 270°, find the value of cosθ.
A
\(\frac{4}{5}\)
B
\(\frac{3}{5}\)
C
-\(\frac{4}{5}\)
D
-\(\frac{3}{5}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
tan\theta = \(frac{3}{4}\) → tan\theta = 0.75
\theta = tan\(^{-1}\)[0.75] → 36.8698^\circ
cos\theta = cos[36.8698^\circ]
→ 0.800 or \(frac{4}{5}\)
However; in the third quadrant Cosine is negative
i.e -\(frac{4}{5}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61240)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, △XYZ is produced to T. if |XY| = |ZY| and ∠XYT = 40°, find ∠XZT
Attached Figure / Diagram
A
110°
B
130°
C
140°
D
180°
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Isosceles Triangle. Two equal sides. Two equal angles ;
∠ X = ∠ Z
∠Y + ∠X + Z = 180 --> 40^\circ + ∠X + ∠X = 180^\circ
∠2X = 180^\circ - 40^\circ = 140^\circ
∠X = 70^\circ, i.e ∠X = 70 and ∠Z = 70^\circ.
The exterior angle of triangle is the sum of two opposite internal angles
: ∠X + ∠Y = ∠XZT
=70^\circ + 40^\circ = 110^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61273)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
A trapezium of parellel sides 10cm and 21cm and height 8cm is inscribed in a circle of radius 7cm. calculate the area of the region not covered by the trapezium.
π =\(\frac{22}{7}\)
A
84cm\(^2\)
B
80cm\(^2\)
C
30cm\(^2\)
D
94cm\(^2\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of circle = \pir\(^2\) = \(\frac{22}{7}\) * 7\(^2\) = 154cm\(^2\)
Area o Trapezium = \(\frac{(a + b) h}{2}\) = \(\frac{(10 + 21)8}{2}\) = 124cm\(^2\)
Area of the region not covered = Area of (Circle - Trapezium)
= 154 - 124
= 30cm\(^2\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61356)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
find the value of (x+y)
Attached Figure / Diagram
A
215°
B
70°
C
135°
D
145°
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
sum of angle in a straight line = 180^\circ
35 + m = 180
m = 180 - 35 = 145^\circ
Corresponding angles are congruent i.e equal to each other
Angle X and M are congruent = 145
sum of angle in a straight line = 180^\circ
110 + n = 180
m = 180 - 110 = 70^\circ
Angle Y and N are congruent = 70^\circ
\(\therefore\) ( x + y) = ( 145 + 70)^\circ
(x + y) = 215^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61422)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, \(\overline{MP}\) is a tangent to the circle NQR, ∠NQR, ∠PNQ = 64 and | \(\overline{RQ}\) | = | \(\overline{RN}\) |. Find the angle market t.
Attached Figure / Diagram
A
130°
B
115°
C
58°
D
68°
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Alternate segments are equal : R^\circ = 64^\circ
An isosceles triangle has two angles ( Q^\circ & N^\circ ) equal as A^\circ
R^\circ + A^\circ + A^\circ = 180^\circ
2A^\circ= 180 - 64
2A^\circ = 116^\circ
A = \(\frac{116}{2}\)
A = 58.
: Q^\circ = 58, N^\circ = 58 and T^\circ = 58
As Alternate segments are equal: T = Q
Source: WAEC / JAMB / NECO Classroom Archive (Record #61423)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQRS is a circle. find the value of x.
Attached Figure / Diagram
A
50°
B
30°
C
80°
D
100°
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Opp. angles in a cyclic quadrilateral always add up to 180^\circ
∠P + ∠R & ∠Q + ∠S = 180
x + x+y = 180
2x + y = 180... i
2y - 30 + x = 180
2y + x = 180 + 30
x + 2y = 210 ... ii
Elimination method:
(2x + y = 180) * 1 --> 2x + y = 180
(x + 2y = 210) * 2 --> 2x + 4y = 420
Subtracting both equations
- 3y = - 240
y = 80^\circ
using eqn i
2x + y = 180
2x + 80 = 180
2x = 100
x = 50^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61425)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
Given that sin x = \(\frac{3}{5}\), 0 ≤ x ≤ 90, evaluate (tan x + 2cosx)
A
2\(\frac{11}{20}\)
B
2\(\frac{7}{20}\)
C
\(\frac{11}{20}\)
D
\(\frac{1}{20}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Sin x = \(\frac{opp}{hyp}\)
sinx = \(\frac{3}{5}\)
using Pythagoras' theorem
hyp\(^2\) = opp\(^2\) + adj\(^2\)
adj\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9
adj\(^2\) = 16
adj = \sqrt 16
adj = 4.
tanx = \(\frac{opp}{adj}\)
= \(\frac{3}{4}\)
cosx = \(\frac{adj}{hyp}\)
= \(\frac{4}{5}\)
(tanx + 2cosx) = \(\frac{3}{4}\) + 2(\(\frac{4}{5}\))
= \(\frac{15 + 32}{20}\)
= \(\frac{47}{20}\) or
2 \(\frac{7}{20}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61427)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, line \(\overline{EC}\) is a diameter of the circle ABCDE.
If angle ABC equals 158°, find ∠ADE
Attached Figure / Diagram
A
112
B
90
C
68
D
22
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the given diagram;
Source: WAEC / JAMB / NECO Classroom Archive (Record #61429)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQRS is a circle. \(\overline{|PQ|}\) = \(\overline{|QS|}\). ∠SPR = 26° and the interior angles of PQS are in the ratio 2:3 :3.
Calculate: (i) PQR; (ii) RPQ; (iii) PRQ
(b) The coordinates of two points P and Q in a plane are (7, 3) and (5, x) respectively, where X is a real number.
If |PQ| = 29units, find the value of x.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) ∠PSQ = ∠QPS
∠PQS = \(\frac{2}{8}\) x 180^\circ
=45^\circ
∠SPQ = ∠PSQ
= \(\frac{3}{8}\) x 180 = 67.5^\circ
∠SQR = ∠SPR= 26^\circ
∠PQR = ∠PQS + ∠SQR
= 45^\circ+ 26^\circ
= 71^\circ
(ii) ∠RPQ = ∠SPQ - ∠SPR
= 67.5^\circ- 26^\circ
= 41.5^\circ
(iii) ∠PRQ = 180^\circ - 41.5^\circ - 71^\circ
= 67.5^\circ
(b) |PQ| = \(\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}\)
29 = \(\sqrt{ (5 - 7)^2 + (x - 3)^2}\)
Take the square of both sides
\(29^2\) = \(-2)^2 + (x - 3)^2
841 = 4 + x\(^2\) - 6x + 9
841 = x\(^2\) - 6x + 13
0 = x\(^2\) - 6x + 13 - 841
x\(^2\) - 6x - 828 = 0
Using the formula method
where a = 1, b = -6, and c = -828
x = 31.93095 or - 25.93095
Therefore x ≈ 31. 93 or x = - 25.93
Source: WAEC / JAMB / NECO Classroom Archive (Record #61563)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Diagram
Solution
(a) On Sam's first birthday celebration, his grandfather deposited an amount of S 1,000.00 in a bank compound at 4 % interest annually.
Find how much is in the account if Sam is 4 years old.
(b) In the diagram above, ABCD are points on the circle centre O. If |AB| = |BC| and ∠ADC= 50°, find ∠BAD.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Amount at end of the 2nd year = \(\frac{104}{100}\) x $1000
= $1,040.00
Amount at end of 3rd year = \(\frac{104}{100}\) x $1040
= $1,081.60
Amount at end of 4th year = \(\frac{104}{100}\) x $1081.6
= $1,124.86
(b) ∠ADC + ∠DCA+ ∠CAD = 180^\circ
50^\circ + 90^\circ+ ∠CAD = 180^\circ
∠CAD = (180 - 140)^\circ
= 40
∠ADC + ∠ABC = 180^\circ
50^\circ + ∠ABC = 180^\circ
∠ABC =130^\circ
∠BAC+ ∠BCA + ∠ABC = 180^\circ
But ∠BAC = ∠BCA
2∠BAC+ ∠130^\circ = 180^\circ
2 ∠BAC=50^\circ
∠BAC = 25^\circ
∠BAD = ∠CAD + ∠BAC
= 40^\circ+25^\circ = 65^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #61564)
Mathematics
UTME / JAMB 2021
Geometry and Trigonometry
Solution
determine the maximum value of y=3x\(^2\) + 5x - 3
A
6
B
0
C
2
D
No correct option
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y=3x\(^2\) + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = \(\frac{-5}{6}\)
Maximum value: 3 \( ^2{\frac{-5}{6}}\) + 5 \(\frac{-5}{6}\) - 3
3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3
Using the L.C.M. 36
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
No correct option
Source: Myschool Past Questions Archive (Record #60628)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
The height of an equilateral triangle of side is 10\(\sqrt{3}\)cm. Calculate its perimeter.
A
20cm
B
60cm
C
40cm
D
30cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Height of an equilateral triangle, h = a\(\frac{\sqrt{3}}{2}\), where a is the side of the equilateral triangle.
10\(\sqrt{3}\) = a\(\frac{\sqrt{3}}{2}\)
cross multiply--> 2 x 10\(\sqrt{3}\) = a\(\sqrt{3}\)
\(\sqrt{3}\) cancels out \(\sqrt{3}\) on both sides
20 = a
The perimeter of an equilateral triangle is: P = 3a
P = 3 * 20 = 60cm
Source: Myschool Past Questions Archive (Record #61207)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
A man left town am at 10:00 AM and traveled by car to town N at an average speed of 72 km/h.
He spent 2hours for a meeting and returned through town M by bus at an average speed of 40KM/H.
If the distance covered by the bus was 2km longer than that of the car and he arrived at town M at 1 :55PM.
calculate distance from M to N.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Time spent traveling = 3 hours 55 minutes - 2 hours
= 1hr 55mins
Let x be the distance from M - N
and t , time spent from M - N
Then t = x / 72
Return time = (23 / 12) - t
(23 / 12) - t = (x + 2) / 40
(23 / 12) - ( x / 72) = (x + 2) / 40
\(\frac{138 - x}{9}\) = \(\frac{x + 2}{5}\)
690 - 5x = 9x + 18
14x = 672
x = 48km
Source: Myschool Past Questions Archive (Record #61469)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270".
Calculate, correct: (a) two significant figures, |YZ|
(b) The nearest degree, the bearing of Y from Z
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
|YZ|\(^2\) = 15\(^2\) + 20\(^2\)
= 225 + 400 = 625
YZ = \sqrt625
YZ = 25
(b) tan \theta = \(\frac{15}{20}\)
tan \theta = 0.75
\theta = tan -1 (0.75)
= 36.87^\circ
Bearing of Y from Z = (90 + 36.87)^\circ = 126.87^\circ
= 127^\circ
Source: Myschool Past Questions Archive (Record #61470)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
The table shows the distribution of the number of hours per day spent in studying by 50 students.
Number of hours per day
4
5
6
7
8
9
10
11
Number of students
5
7
5
9
12
4
3
5
Calculate, correct to two decimal places,
the: (a) mean; (b) standard deviation.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
x
f
fx
fx\(^2\)
4
5
20
80
5
7
35
175
6
5
30
180
7
9
63
441
8
12
96
768
9
4
36
324
10
3
30
300
11
5
55
605
£f = 50
£fx = 365
£fx\(^2\) = 2,873
(a) Mean(\(\overline{x}\)) = \(\frac{365}{50}\) = 7.30 (2 d.p.)
(b) Standard deviation = \sqrt(\(\frac{2873}{50}\)) - (7.3)\(^2\)
= \sqrt4.17
= 2.0421
= 2.04 (2 d.p.)
Source: Myschool Past Questions Archive (Record #61562)