Mathematics (Core), 1992, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

1992

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 48 Questions

#	Question	Ans
21.	In the diagram above, TR = TS and ∠TRS = 60^o. Find the value of x A. 30 B. 45 C. 60 D. 120 E. 150 Show Content Detailed Solution ∠TRS = ∠RST = 60^o ∠PSR + ∠PQR = 180^o, 120^o + x^o = 180^o x = 60^o	C
22.	In the diagram above, O is the center of the circle PQRS and ∠QPS = 36°. Find ∠QOS A. 36^o B. 72^o C. 108^o D. 144^o E. 288^o Show Content Detailed Solution < QOS = 2 < QPS (angle subtended at the centre) \(\therefore\) < QOS = 2 x 36° = 72°	B
23.	In the diagram above, O is the center of the circle QOR is a diameter and ∠PSR is 37°. Find ∠PRQ A. 37^o B. 53^o C. 65^o D. 127^o E. 147^o Show Content Detailed Solution Construction: Join PQ. Then < RSP = 37° = < RQP (angles on the same segment) But < RPQ = 90° (angle in a semi-circle) \(\therefore\) < PRQ = 180° - (90° + 37°) = 53°	B
24.	In the diagram above, ML//PQ and NP//QR, if ∠LMN = 40° and ∠MNP = 55°. Find ∠QPR A. 15^o B. 25^o C. 35^o D. 40^o E. 55^o Show Content Detailed Solution Draw line ANB// ML < MNA = 40° (alternate angle AB/ ML) < ANP = 55° - 40° = 15° < POR = 15° = < ANP (corresponding angles, PN // RQ)	A
25.	The angles marked in the diagram above are measured in degrees. Find x A. 15^o B. 24^o C. 30^o D. 36^o E. 48^o Show Content Detailed Solution Sum of exterior angles = 360° 2x + 3x + 2x + 3x + 5x = 360° 15x = 360° x = 360°/15 = 24°	B
26.	Find the value of t in the diagram above A. 5.6cm B. 6.5cm C. 6.6cm D. 6.8cm E. 7.0cm Show Content Detailed Solution t/4 + t = 7/7 + 5 t = 5.6	A
27.	A regular polygon has each of its exterior angles as 18^o. How many sides has the polygon A. 10 B. 11 C. 12 D. 20 E. 24 Show Content Detailed Solution ³⁶⁰/₁₈ = 20	D
28.	In the diagram above, PQT is an isosceles triangle.\|PQ\| = \|QT\|, ∠SRQ = 75°, ∠QPT = 25° and PQR is straight line. Find ∠RST A. 20^o B. 50^o C. 55^o D. 70^o E. 75^o Show Content Detailed Solution < PTQ = 25° (base angles of an isos. triangle) \(\therefore\) < PQT = 180° - (25° + 25°) = 130° (sum of angles in triangle PQT) \(\therefore\) < RST = 130° - 75° = 55° (exterior angle = sum of 2 opp. interior angles)	C
29.	Sin 60° has the same value as I. Sin 120° II. cos 240° III. -sin 150° IV. cos 210° V. sin 240° A. I only B. II only C. IV only D. III only E. IV and V only Show Content Detailed Solution In the second quadrant, \(\sin 120 = \sin (180 - 120)\) = \(\sin 60\)	A
30.	If cos θ = 5/13, what is the value of tan \(\theta\) for 0 < θ < 90° ? A. 13 B. 5 C. 13/5 D. 12/5 E. 5/12 Show Content Detailed Solution \(\cos \theta = \frac{5}{13}\) \(\implies\) In the right- angled triangle, with an angle \(\theta\), the adjacent side to \(\theta\) = 5 and the hypotenuse = 13. \(\therefore 13^2 = opp^2 + 5^2\) \(opp^2 = 169 - 25 = 144 \implies opp = \sqrt{144}\) = 12. \(\tan \theta = \frac{opp}{adj} = \frac{12}{5}\)	D

21.	In the diagram above, TR = TS and ∠TRS = 60^o. Find the value of x A. 30 B. 45 C. 60 D. 120 E. 150 Show Content Detailed Solution ∠TRS = ∠RST = 60^o ∠PSR + ∠PQR = 180^o, 120^o + x^o = 180^o x = 60^o	C
22.	In the diagram above, O is the center of the circle PQRS and ∠QPS = 36°. Find ∠QOS A. 36^o B. 72^o C. 108^o D. 144^o E. 288^o Show Content Detailed Solution < QOS = 2 < QPS (angle subtended at the centre) \(\therefore\) < QOS = 2 x 36° = 72°	B
23.	In the diagram above, O is the center of the circle QOR is a diameter and ∠PSR is 37°. Find ∠PRQ A. 37^o B. 53^o C. 65^o D. 127^o E. 147^o Show Content Detailed Solution Construction: Join PQ. Then < RSP = 37° = < RQP (angles on the same segment) But < RPQ = 90° (angle in a semi-circle) \(\therefore\) < PRQ = 180° - (90° + 37°) = 53°	B
24.	In the diagram above, ML//PQ and NP//QR, if ∠LMN = 40° and ∠MNP = 55°. Find ∠QPR A. 15^o B. 25^o C. 35^o D. 40^o E. 55^o Show Content Detailed Solution Draw line ANB// ML < MNA = 40° (alternate angle AB/ ML) < ANP = 55° - 40° = 15° < POR = 15° = < ANP (corresponding angles, PN // RQ)	A
25.	The angles marked in the diagram above are measured in degrees. Find x A. 15^o B. 24^o C. 30^o D. 36^o E. 48^o Show Content Detailed Solution Sum of exterior angles = 360° 2x + 3x + 2x + 3x + 5x = 360° 15x = 360° x = 360°/15 = 24°	B

26.	Find the value of t in the diagram above A. 5.6cm B. 6.5cm C. 6.6cm D. 6.8cm E. 7.0cm Show Content Detailed Solution t/4 + t = 7/7 + 5 t = 5.6	A
27.	A regular polygon has each of its exterior angles as 18^o. How many sides has the polygon A. 10 B. 11 C. 12 D. 20 E. 24 Show Content Detailed Solution ³⁶⁰/₁₈ = 20	D
28.	In the diagram above, PQT is an isosceles triangle.\|PQ\| = \|QT\|, ∠SRQ = 75°, ∠QPT = 25° and PQR is straight line. Find ∠RST A. 20^o B. 50^o C. 55^o D. 70^o E. 75^o Show Content Detailed Solution < PTQ = 25° (base angles of an isos. triangle) \(\therefore\) < PQT = 180° - (25° + 25°) = 130° (sum of angles in triangle PQT) \(\therefore\) < RST = 130° - 75° = 55° (exterior angle = sum of 2 opp. interior angles)	C
29.	Sin 60° has the same value as I. Sin 120° II. cos 240° III. -sin 150° IV. cos 210° V. sin 240° A. I only B. II only C. IV only D. III only E. IV and V only Show Content Detailed Solution In the second quadrant, \(\sin 120 = \sin (180 - 120)\) = \(\sin 60\)	A
30.	If cos θ = 5/13, what is the value of tan \(\theta\) for 0 < θ < 90° ? A. 13 B. 5 C. 13/5 D. 12/5 E. 5/12 Show Content Detailed Solution \(\cos \theta = \frac{5}{13}\) \(\implies\) In the right- angled triangle, with an angle \(\theta\), the adjacent side to \(\theta\) = 5 and the hypotenuse = 13. \(\therefore 13^2 = opp^2 + 5^2\) \(opp^2 = 169 - 25 = 144 \implies opp = \sqrt{144}\) = 12. \(\tan \theta = \frac{opp}{adj} = \frac{12}{5}\)	D