Year : 
1995
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 48 Questions

# Question Ans
31.

In the diagram above; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35°, find ∠BAO.

A. 20o

B. 25o

C. 30o

D. 35o

E. 40o

Detailed Solution

< DBC = 35° (base angles of an isosceles triangle)
< CDB = 180° - (35° + 35°)
= 110°
< ADB = 70°; < ADB = 90°
\(\therefore\) < BAO = 180° - (70° + 90°)
= 20°
32.

In the diagram above, AO is perpendicular to OB. Find x

A. 7.5o

B. 15o

C. 22.5o

D. 30o

E. 38.6o

Detailed Solution

4x + 3x + 2x + 90 =360° [angle at a point]
9x + 90 = 360°
9x = 360° - 90°
9x = 270
x = 270/9
x = 30°
33.

In the diagram above, PQ is parallel to TU, ∠PQR = 50°, ∠QRS = 86° and ∠STU = 64°. Calculate the value of x.

A. 136o

B. 120o

C. 108o

D. 100o

E. 96o

Detailed Solution

y = 86° - 50° [Alternate ∠s are equal]
y = 86° - 50° = 36°
a = y = 36° [Alternate ∠s]; b = 64° [Alternate ∠s;]
x = a + b = (36° + 64°)
=100°
34.

If log\(_{10}\) x = \(\bar{2}.3675\) and log\(_{10}\) y = \(\bar{2}.9738\), what is the value of x + y, correct lo three significant figures?

A. O.117

B. 0.118

C. 0.903

D. O.944

E. 0.946

Detailed Solution

log\(_{10}\) x = \(\bar{2}.3675\) => x = 0.023
log\(_{10}\) y = \(\bar{2}.9738\) => y = 0.094
x + y = 0.117
35.

In the diagram above, AB//CD, the bisector of ∠BAC and ∠ACD meet at E. Find the value of ∠AEC

A. 30o

B. 45o

C. 60o

D. 75o

E. 90o

Detailed Solution

Ae is a bisector of BAC => EAC = 45o
CE is the bisector of ACD => ACE = 45o
AEC = 180o - [EAC + ACE]
    = 180o - (45o + 45o) = 180o - 90o = 90o
36.

In the diagram above, AB//CD. What is the size of the angle marked x?

A. 103o

B. 93o

C. 77o

D. 62o

E. 52o

Detailed Solution

a = 52o [vertical opp. ∠s are equal]
a = b = 52o [Alternate ∠s]
b = c = 52o [vertical opp ∠s]
c + x + 35o = 180o [sum of angles in a triangle]
52o + x + 35o = 180o
x + 87o = 180o
x = 180o - 87o = 93o <
37.

The locus of a point which is equidistant from two given fixed points is the

A. perpendicular bisector of the straight line joining them

B. angle bisector of the straight lines joining the points to the origin

C. perpendiculars to the straight line joining them

D. parallel-line to the straight line joining them

E. a line making an acute angle with the line joining the two points

A

38.

In the diagram above, the value of angles b + c is

A. 180o

B. 90o

C. 45o

D. ao

E. do

Detailed Solution

a + c + d - 180o[sum of angles in triangle]
x = a[vertical opp. ∠s]
y + x + d = 180o[sum of ∠s in a Δ]
y + a + d = 180o
y + a + d = a + c + d
y = c
but b + y = 180o [sum of ∠s on a straight line ]
b + c = 180o; Ans = A = 180o
39.

Which of the following angles is an exterior angle of a regular polygon?

A. 95o

B. 85o

C. 78o

D. 75o

E. 72o

Detailed Solution

The formula for the exterior angle of a regular polygon = \(\frac{360}{n}\).
Among the options, only 72° is divisible by 360° to give an integer.
40.

In ΔABC above, BC is produced to D, /AB/ = /AC/ and ∠BAC = 50o. Find ∠ACD

A. 50o

B. 60o

C. 65o

D. 100o

E. 115o

Detailed Solution

ACB = ABC [Base angles of isoceles triangle]
ACB + ABC + 50o = 180o[sum of angles in a triangle]
ACB + ABC = 180o - 50o = 130o
ACB
130o/2
= 65o

ACD + ACB = 180o[sum of ∠s on a straight line]
ACD + 65o = 180o
ACD =
31.

In the diagram above; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35°, find ∠BAO.

A. 20o

B. 25o

C. 30o

D. 35o

E. 40o

Detailed Solution

< DBC = 35° (base angles of an isosceles triangle)
< CDB = 180° - (35° + 35°)
= 110°
< ADB = 70°; < ADB = 90°
\(\therefore\) < BAO = 180° - (70° + 90°)
= 20°
32.

In the diagram above, AO is perpendicular to OB. Find x

A. 7.5o

B. 15o

C. 22.5o

D. 30o

E. 38.6o

Detailed Solution

4x + 3x + 2x + 90 =360° [angle at a point]
9x + 90 = 360°
9x = 360° - 90°
9x = 270
x = 270/9
x = 30°
33.

In the diagram above, PQ is parallel to TU, ∠PQR = 50°, ∠QRS = 86° and ∠STU = 64°. Calculate the value of x.

A. 136o

B. 120o

C. 108o

D. 100o

E. 96o

Detailed Solution

y = 86° - 50° [Alternate ∠s are equal]
y = 86° - 50° = 36°
a = y = 36° [Alternate ∠s]; b = 64° [Alternate ∠s;]
x = a + b = (36° + 64°)
=100°
34.

If log\(_{10}\) x = \(\bar{2}.3675\) and log\(_{10}\) y = \(\bar{2}.9738\), what is the value of x + y, correct lo three significant figures?

A. O.117

B. 0.118

C. 0.903

D. O.944

E. 0.946

Detailed Solution

log\(_{10}\) x = \(\bar{2}.3675\) => x = 0.023
log\(_{10}\) y = \(\bar{2}.9738\) => y = 0.094
x + y = 0.117
35.

In the diagram above, AB//CD, the bisector of ∠BAC and ∠ACD meet at E. Find the value of ∠AEC

A. 30o

B. 45o

C. 60o

D. 75o

E. 90o

Detailed Solution

Ae is a bisector of BAC => EAC = 45o
CE is the bisector of ACD => ACE = 45o
AEC = 180o - [EAC + ACE]
    = 180o - (45o + 45o) = 180o - 90o = 90o
36.

In the diagram above, AB//CD. What is the size of the angle marked x?

A. 103o

B. 93o

C. 77o

D. 62o

E. 52o

Detailed Solution

a = 52o [vertical opp. ∠s are equal]
a = b = 52o [Alternate ∠s]
b = c = 52o [vertical opp ∠s]
c + x + 35o = 180o [sum of angles in a triangle]
52o + x + 35o = 180o
x + 87o = 180o
x = 180o - 87o = 93o <
37.

The locus of a point which is equidistant from two given fixed points is the

A. perpendicular bisector of the straight line joining them

B. angle bisector of the straight lines joining the points to the origin

C. perpendiculars to the straight line joining them

D. parallel-line to the straight line joining them

E. a line making an acute angle with the line joining the two points

A

38.

In the diagram above, the value of angles b + c is

A. 180o

B. 90o

C. 45o

D. ao

E. do

Detailed Solution

a + c + d - 180o[sum of angles in triangle]
x = a[vertical opp. ∠s]
y + x + d = 180o[sum of ∠s in a Δ]
y + a + d = 180o
y + a + d = a + c + d
y = c
but b + y = 180o [sum of ∠s on a straight line ]
b + c = 180o; Ans = A = 180o
39.

Which of the following angles is an exterior angle of a regular polygon?

A. 95o

B. 85o

C. 78o

D. 75o

E. 72o

Detailed Solution

The formula for the exterior angle of a regular polygon = \(\frac{360}{n}\).
Among the options, only 72° is divisible by 360° to give an integer.
40.

In ΔABC above, BC is produced to D, /AB/ = /AC/ and ∠BAC = 50o. Find ∠ACD

A. 50o

B. 60o

C. 65o

D. 100o

E. 115o

Detailed Solution

ACB = ABC [Base angles of isoceles triangle]
ACB + ABC + 50o = 180o[sum of angles in a triangle]
ACB + ABC = 180o - 50o = 130o
ACB
130o/2
= 65o

ACD + ACB = 180o[sum of ∠s on a straight line]
ACD + 65o = 180o
ACD =