Year : 
1979
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 50 of 51 Questions

# Question Ans
41.

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

Detailed 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
42.

In the figure, \(\bigtriangleup\) ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and o then AE is equal to

A. 3\(\sqrt{2}\)

B. 2\(\sqrt{3}\)

C. \(\frac{\sqrt{3}}{2}\)

D. \(\frac{2}{\sqrt{3}}\)

Detailed 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 = \(\frac{\sqrt{3}}{2}\)<
43.

In this figure, PQRS is a parallelogram, PS = PT and < PST = 55\(^o\). The size of <PQR is

A. 125o

B. 120o

C. 115o

D. 110o

E. 10o

Detailed 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\(^o\)
44.

If O is the centre of the circle, < POS equls

A. 70o

B. 75o

C. 105o

D. 140o

E. 150o

Detailed Solution

Since O is the centre of the circle < POS = 150o

i.e. < substended at the centre is twice that substended at any part of the circumference
45.

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

  • A. 70o
  • B. 90o
  • C. 80o
  • D.<

A. 70o

B. 90o

C. 80o

D. 40o

E. 60o

Detailed Solution

The site of POQ = 80o
46.

In the figure, PQ is parallel to SQ ; QS bisets < PSQ, < PQS is 65o and < RPS is 20o. What is the size of < PRS?

A. 45o

B. 35o

C. 40o

D. 30o

E. 42o

Detailed Solution

PSR = 65o x 2

= 130o

PRS = 180o - (130o + 20o)

= 30o
47.

(Numbers indicate the lengths of the sides of the triangles) If the area of \(\bigtriangleup\) PQR is k2sq. units what is the area of the shades portion?

A. units

B. units

C. units

D. units

E. u

Detailed Solution

Area of shaded portion = Area of triangle PQR - Area of inner triangle
Area of triangle given 3 sides a, b, c = \(\sqrt{s(s - a)(s - b)(s - c)}\)
where \(s = \frac{a + b + c}{2} \)
Area of PQR :
\(s = \frac{3 + 5 + 6}{2} = \frac{14}{2} = 7\)
Area = \(\sqrt{7(7 - 3)(7 - 5)(7 - 6)}\)
= \(\sqrt{7(4)(2)(1)} = \sqrt{56}\)
\(\implies K^{2} = \sqrt{56}\)
Area of inner triangle :
\(s = \frac{2 + 4 + \frac{10}{3}}{2} = \frac{14}{3}\)
Area = \(\sqrt{\frac{14}{3} (\frac{14}{3} - 2)(\frac{14}{3} - 4)(\frac{14}{3} - \frac{10}{3})}\)
= \(\sqrt{\frac{14}{3} (\frac{8}{3})(\frac{2}{3})(\frac{4}{3})}\)
= \(\sqrt{\frac{896}{81}}\)
= \(\sqrt{\fr
48.

In the parallelogram PQRS, PE is perpendicular to QR. Find the area of the parallelogram.

A. 60cm2

B. 65cm2

C. 72cm2

D. 132cm2

E. 156cm2

Detailed Solution

By Pythagoras, PE\(^2\) = 12\(^2\) - 5\(^2\)

= 144 - 25 = 119

h = PE\(^2\) = √119 = 10.9 ≈ 11cm,

Area of 11gm = b x h

QR = b =(5 + 7)cm = 12cm

area = 12 x 11

= 132cm\(^2\)
49.

PQ is parallel to RS. Calculate the value of x.

A. 20o

B. 40o

C. 60o

D. 80o

E. 100o

Detailed Solution

< D = 180o - 100v

= 80o (< on a str. line)

< s = 60o - alternate angle

x = 180o - (80o + 60o)

180o - 140o = 40o
50.

Find x in the diagram below.

A. 3\(\sqrt{3}\)

B. \(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)

C. \(\frac{(\sqrt{3} - 1)}{\sqrt{3} + 1}\)

D. \(\frac{3 - 1}{\sqrt{3} + 1}\)

Detailed Solution

\(\frac{x + 3}{sin 60^o}\) = \(\frac{x}{sin 306o}\)

3sin 30o = x sin 60o - x sin 30o

= x(sin 60o - sin 30o)

but sin 30o = \(\frac{1}{2}\)

sin 60o = \(\frac{3}{2}\) = \(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)
41.

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

Detailed 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
42.

In the figure, \(\bigtriangleup\) ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and o then AE is equal to

A. 3\(\sqrt{2}\)

B. 2\(\sqrt{3}\)

C. \(\frac{\sqrt{3}}{2}\)

D. \(\frac{2}{\sqrt{3}}\)

Detailed 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 = \(\frac{\sqrt{3}}{2}\)<
43.

In this figure, PQRS is a parallelogram, PS = PT and < PST = 55\(^o\). The size of <PQR is

A. 125o

B. 120o

C. 115o

D. 110o

E. 10o

Detailed 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\(^o\)
44.

If O is the centre of the circle, < POS equls

A. 70o

B. 75o

C. 105o

D. 140o

E. 150o

Detailed Solution

Since O is the centre of the circle < POS = 150o

i.e. < substended at the centre is twice that substended at any part of the circumference
45.

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

  • A. 70o
  • B. 90o
  • C. 80o
  • D.<

A. 70o

B. 90o

C. 80o

D. 40o

E. 60o

Detailed Solution

The site of POQ = 80o
46.

In the figure, PQ is parallel to SQ ; QS bisets < PSQ, < PQS is 65o and < RPS is 20o. What is the size of < PRS?

A. 45o

B. 35o

C. 40o

D. 30o

E. 42o

Detailed Solution

PSR = 65o x 2

= 130o

PRS = 180o - (130o + 20o)

= 30o
47.

(Numbers indicate the lengths of the sides of the triangles) If the area of \(\bigtriangleup\) PQR is k2sq. units what is the area of the shades portion?

A. units

B. units

C. units

D. units

E. u

Detailed Solution

Area of shaded portion = Area of triangle PQR - Area of inner triangle
Area of triangle given 3 sides a, b, c = \(\sqrt{s(s - a)(s - b)(s - c)}\)
where \(s = \frac{a + b + c}{2} \)
Area of PQR :
\(s = \frac{3 + 5 + 6}{2} = \frac{14}{2} = 7\)
Area = \(\sqrt{7(7 - 3)(7 - 5)(7 - 6)}\)
= \(\sqrt{7(4)(2)(1)} = \sqrt{56}\)
\(\implies K^{2} = \sqrt{56}\)
Area of inner triangle :
\(s = \frac{2 + 4 + \frac{10}{3}}{2} = \frac{14}{3}\)
Area = \(\sqrt{\frac{14}{3} (\frac{14}{3} - 2)(\frac{14}{3} - 4)(\frac{14}{3} - \frac{10}{3})}\)
= \(\sqrt{\frac{14}{3} (\frac{8}{3})(\frac{2}{3})(\frac{4}{3})}\)
= \(\sqrt{\frac{896}{81}}\)
= \(\sqrt{\fr
48.

In the parallelogram PQRS, PE is perpendicular to QR. Find the area of the parallelogram.

A. 60cm2

B. 65cm2

C. 72cm2

D. 132cm2

E. 156cm2

Detailed Solution

By Pythagoras, PE\(^2\) = 12\(^2\) - 5\(^2\)

= 144 - 25 = 119

h = PE\(^2\) = √119 = 10.9 ≈ 11cm,

Area of 11gm = b x h

QR = b =(5 + 7)cm = 12cm

area = 12 x 11

= 132cm\(^2\)
49.

PQ is parallel to RS. Calculate the value of x.

A. 20o

B. 40o

C. 60o

D. 80o

E. 100o

Detailed Solution

< D = 180o - 100v

= 80o (< on a str. line)

< s = 60o - alternate angle

x = 180o - (80o + 60o)

180o - 140o = 40o
50.

Find x in the diagram below.

A. 3\(\sqrt{3}\)

B. \(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)

C. \(\frac{(\sqrt{3} - 1)}{\sqrt{3} + 1}\)

D. \(\frac{3 - 1}{\sqrt{3} + 1}\)

Detailed Solution

\(\frac{x + 3}{sin 60^o}\) = \(\frac{x}{sin 306o}\)

3sin 30o = x sin 60o - x sin 30o

= x(sin 60o - sin 30o)

but sin 30o = \(\frac{1}{2}\)

sin 60o = \(\frac{3}{2}\) = \(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)