JAMB UTME - Mathematics - 1987

Question 1 Report

The sine, cosine and tangent of 210o are respectively

\frac{1}{2}

,

\frac{\sqrt{3}}{2}

,

\frac{\sqrt{3}}{2}

\frac{1}{2}

,

\frac{\sqrt{3}}{2}

,

\frac{\sqrt{3}}{3}

\frac{1}{2}

,

\frac{\sqrt{3}}{2}

,

\frac{\sqrt{3}}{2}

$\frac{- 1}{2}$ , $\frac{\sqrt{- 3}}{2}$ , $\frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}{3}$

Answer Details

210o = 180o - 210o = 30o
From ratio of sides, sin -30o = -

\frac{1}{2}

Cos 210o = 180o - 210o = -30o
= cos -30o =

\frac{- 3}{2}

But tan 30o =

\frac{1}{3}

, rationalizing this
=

\frac{1}{3}

x

\frac{3}{3}

=

\frac{3}{3}

∴ =

\frac{- 1}{2}

,

\frac{\sqrt{- 3}}{2}

,

\frac{\sqrt{3}}{3}

Read lesson note on Sine, Cosine And Tangent Of An Angle (WAEC)

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Read lesson note on Coordinate Geometry (WAEC)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Coordinate Geometry

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Question 2 Report

What is the locus or the mid-point of all the chords of length 6cm with circle of radius 5cm and with center 0?

A circle of radius 4 cm and the center 0

The perpendicular bisector of the chords

A straight line passing through 0

A circle of radius 6 cm and with center 0

Answer Details

x2 + 33 = 52
x2 + 9 = 25
x2 = 25 – 9
x2 = 16

x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0

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Circles

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Question 3 Report

Make $y$ the subject of the formula $Z = x^2 + \frac{1}{y^3}$

y =

\frac{1}{(z - x^{2})^{3}}

y =

\frac{1}{(z + x^{2}) \frac{1}{3}}

y =

\frac{1}{(z x^{2}) \frac{1}{3}}

y =

3 \sqrt{z - x^{2}}

Answer Details

Z = x2 +

\frac{1}{y^{3}}

Z - x2 =

\frac{1}{y^{3}}

y3 =

\frac{1}{Z x^{2}}

y =

3 \sqrt{\frac{1}{2} x^{2}}

∴ y =

3 \sqrt{Z - x^{2}}

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Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

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Question 4 Report

Reduce each number to two significant figures and then evaluate $ \frac{0.021741 \times 1.2047}{0.023789} $

0.8

0.9

1.1

1.2

Answer Details

\frac{0.021741 \times 1.2047}{0.023789}

=

\frac{0.022 \times 1.2}{0.024}

(to 216)
=

\frac{0.0264}{0.024}

= 1.1

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Fractions, Decimals And Approximations

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Question 5 Report

If $ \tan \theta = \frac{m^2-n^2}{2mn} $ find $ \sec \theta $

m²+n²

(m²−n²)

m²+n²

2mn

mn

2(m²+n²⁾

m²n²

2m²−n²

Answer Details

Tan

θ

=

\frac{m^{2} - n^{2}}{2 m n}

\frac{Opp}{Adj}

by pathagoras theorem
= Hyp2 = Opp2 + Adj2
Hyp2 = (m2 - n2) + (2mn)2
Hyp2 = m4 - 2m2n4 - 4m2 - n2
Hyp2 = m4 + 2m2 + n2n
Hyp2 = (m2 - n2)2
Hyp2 =

\frac{m^{2} + n^{2}}{2 m n}

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Sine, Cosine And Tangent Of An Angle

Angles Of Elevation And Depression

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Question 6 Report

Factorize 62x + 1 + 7(6x) - 5

[3(6x) - 5][2(6x)m + 1]

[3(6x) + 5][2(6x) - 1]

[(6x) - 5][3(6x) + 1]

[2(6x) + 5][3(6x) - 1]

Answer Details

62x + 1 + 7(6x) - 5
Let 6x = y....(i)
6yx + 7y - 5 = 0
(3y + 5)(2y - 1) = 0.....(i)
Subt. for y from eqn. (i) in eqn. (ii)
= [3(6x) + 5][2(6x) - 1]

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Algebraic Expressions

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Question 7 Report

Four boys and ten girls can cut a field in 5 hours if the boys work at $ \frac{5}{4} $ the rate at which the girls work. How many boys will be needed to cut the field in 3 hours?

180

60

25

20

Answer Details

Let x represents number of boys that can work at

\frac{5}{4}

the rate at which the 10 girls work
For 1hr. x boys will work for

\frac{\frac{1}{5}}{4}

x 10
x =

\frac{4}{5}

x 10 = 8 boys
8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs
For 3 hrs,

\frac{12 \times 5}{3}

boys will be needed = 20 boys

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Solution Of Linear Equations

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Question 8 Report

A rectangular lawn has an area of 1815 square yards. if its length is 50 meters, find its width in meters given that 1 metre equals 1.1 yards

39.93

35.00

33.00

30.00

Answer Details

1m = 1.1 yard, length(L)= 50m
= (50 x 1.1)yards
= 55 yards
Area(A) = length(L) y width (W)
1815 = 55y width (W)
width (w) =

\frac{1815}{55}

= 33 years

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Areas

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Question 9 Report

The formula Q = 1.5 + 0.5n gives the cost Q(in Naira)of feeding n people one additional person

350k

200k

150k

50k

Answer Details

Q = 1.5 + 0.5n gives the cost 1(in Naira) of feeding n people for a week. Extra cost of feeding one additional person = n = 1
Subt. for n in the formula Q = 1.5 + 0.5(1)
= 15 + 0.5(2)
Q = ₦2
= 200k

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Change Of Subject Of A Formula/Relation

Areas

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Question 10 Report

A man invests a sum of money at 4% per annum simple interest. After 3 years, the principal amounts to ₦7,000.00. Find the sum invested.

₦7,840.00

₦6,250.00

₦6,160.00

₦5,833.33

Modular Arithmetic

Areas

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Question 11 Report

Find the probability of selecting a figure which is a parallelogram from a square, a rectangle, a rhombus, a kite and a trapezium

3

5

2

5

4

5

1

5

Answer Details

A rectangle, a square and a rhombus are the only parallelogram
∴ Probability of a parallelogram =

\frac{3}{5}

View Answer

Question 12 Report

Find the least length of a rod which can be cut into exactly equal strips, each of either 40cm or 48cm in length

120cm

240cm

360cm

480cm

Answer Details

The least length is

\frac{40}{48}

=

\frac{5}{8}

for the rod to be cut in exactly equal trips
Ratio

\frac{5}{6}

:

\frac{48}{40}

\frac{\frac{5}{6}}{\frac{40}{48}}

= 1

\frac{5}{6}

x

\frac{48}{40}

=

\frac{240}{240}

= 1
The least length = 240cm

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Read lesson note on Number Bases (WAEC)

Positive And Negative Integers, Rational Numbers

Number Bases

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Question 13 Report

Find the eleventh term of the progression 4, 8, 16.....

213

212

211

210

Answer Details

a = 4, r =

\frac{4}{2}

\frac{8}{4}

= 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212

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Sequence And Series

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Question 14 Report

The figure is an example of the construction of a

perpendicular bisector of a given straight line QR

perpendicular from a given point to a given line QR

perpendicular to a line from a given point P on that line

given angle

Answer Details

QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR

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Euclidean Geometry

Simple Operations On Algebraic Expressions

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Question 15 Report

Two brothers invested a total of ₦5,000.00 on a farm project, the farm yield was sold for ₦15,000.00 at the end of the season. If the profit was shared in the ratio 2 : 3, what is the difference in the amount to profit received by the brothers?

₦2,000.00

₦4,000.00

₦6,000.00

₦10,000.00

Answer Details

Total amount invested by A and B = ₦5,000
farm yield was sold for ₦15,000.00
profit = 15,000.00 - 5,000.00
= ₦10,000.00
Profit was shared in ratio 2 : 3
2 + 3 = 5
A received

\frac{2}{5}

of profit =

\frac{2}{5}

x 10,000 = ₦4,000.00
A receive

\frac{3}{5}

of profit =

\frac{3}{5}

x 10,000 = ₦6,000.00
Difference in profit received = ₦6,0000 - ₦4,000.00
= ₦2,000.00

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Question 16 Report

From two points x and y, 8m apart, and in line with a pole, the angle of elevation of the top of the pole are 30o and 60o respectively. Fins the height of the pole assuming that x, y and the foot of the pole are the same horizontal plane and x and y are on the same side of the pole.

\sqrt{6}

4

\sqrt{3}

√3

\frac{12}{\sqrt{3}}

Answer Details

From the diagram, WYZ = 60o, XYW = 180o - 60o
= 120o
LX = 30o
XWY = 180o - 120o + 30o
XWY = 30o
WXY = XYW = 30o
Side XY = YW
YW = 8m, sin 60o =

\frac{3}{2}

? sin 60o =

\frac{h}{Y W}

, sin 60o =

\frac{h}{8}

h = 8 x

\frac{3}{2}

= 4

\sqrt{3}

View Answer

Question 17 Report

If $ P = \frac{2}{3}\left(\frac{1-r^2}{n^2}\right) $, find n when r = $ \frac{1}{3} $ and p = 1

3

2

1

3

2

3

Answer Details

If P =

\frac{2}{3}

(

\frac{1 - r^{2}}{n^{2}}

), find n when r =

\frac{1}{3}

and p = 1
p =

\frac{2 (1 - r^{2})}{3 n^{2}}

when r =

\frac{1}{3}

and p = 1
1 =

\frac{2}{3}

\frac{(1 - (\frac{1}{3})^{2})}{n^{2}}

n2 =

\frac{2 (3 - 1)}{3 \times 3}

n2 =

\frac{2 \times 2}{3 \times 3}

=

\frac{4}{9}

n =

\frac{4}{9}

=

\frac{2}{3}

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Probability

Measures Of Location

Quadratic Equations

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Question 18 Report

Convert 241 in base 5 to base 8,

718

1078

1768

2418

Answer Details

2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o
50 + 20 + 1 = 7110
Convert 7110 to base 8
$\begin{array}{cc} 8 & 71 \\ 8 & 8 R 7 \\ 8 & 1 R 0 \\ 8 & 0 R 1 \end{array}$
= 1078

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Number Bases

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Question 19 Report

Simplify $ \frac{x^2-y^2}{2x^2+xy-y^2} $

$\frac{x + y}{2 x + y}$

$\frac{x + y}{2 x - y}$

$\frac{x - y}{2 x - y}$

$\frac{x - y}{2 x + y}$

Answer Details

$\frac{x^{2} - y^{2}}{2 x^{2} + x y - y^{2}}$

$2 x^{2} + x y - y^{2} = 2 x^{2} - x y + 2 x y - y^{2}$

= $x (2 x - y) + y (2 x - y)$

= $(x + y) (2 x - y)$

$\frac{x^{2} - y^{2}}{2 x^{2} + x y - y^{2}} = \frac{(x + y) (x - y)}{(x + y) (2 x - y)}$

= $\frac{x - y}{2 x - y}$

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Solution Of Linear Equations

Algebraic Expressions

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Question 20 Report

The numbers 3, 2, 8, 5, 7, 12, 9 and 14 are the marks scored by a group of students in a class test. If P is scored by an group of students in a class test. If P is the mean and Q the median, the P + Q is

18

17

\frac{1}{2}

16

15

Answer Details

Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14
Mean = P and median = Q
P =

\frac{2 + 2 + 5 + 7 + 8 + 9 + 12 + 14}{8}

\frac{60}{8}

= 7.5
median (Q) =

\frac{7 + 8}{2}

\frac{15}{2}

= 7.5
P + Q = 7.5 + 7.5 = 15

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Number Bases

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Question 21 Report

By selling 20 oranges for ₦1.35 a trader makes a profit of 8%. What is his percentage gain or loss if he sells the same 20 oranges for ₦1.10?

8%

10%

12%

15%

Answer Details

profit 8% of ₦1.35 =

\frac{8}{100}

x ₦1.35 = ₦0.08
Cost price = ₦1.35 - ₦0.10 = ₦1.25
If he sells the 20 oranges for ₦1.10 now
%loss =

\frac{actual loss}{Cost price}

x 100

\frac{125 - 1.10}{1.25}

x 100
=

\frac{0.15 \times 100}{1.25}

=

\frac{15}{1.25}

= 12%

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Modular Arithmetic

Ratio, Proportions And Rates

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Question 22 Report

Simplify without using tables $ \frac{\log_2 6}{\log_2 8} - \frac{\log_2 3}{\log_2 \frac{1}{2}} $

1

5

1

2

-1

2

log²3

log²7

Answer Details

log26 - log23 = log2 (

\frac{6}{3}

) = log22
log28 - 2

\frac{1}{2}

log28 - log2

\frac{1}{2}

log28 - log2

\frac{1}{4}

\frac{l o g_{2} 2}{l o g_{2} 32} = \frac{l o g_{2} 2}{5 l o g_{2} 2}

=

\frac{1}{5}

N.B. log22 = 1

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Logarithms

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Question 23 Report

An $ (n-2)^2 $ sided figure has n diagonals, find the number n of diagonals for a 25 sided figure

8

7

6

9

10

Variation

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Question 24 Report

Below are the scores of group of students in a music test:

Scores	1	2	3	4	5	6	7	8	9
No of students	3	6	10	8	6	5	2	4	12

If CF(x) is the number of students with scores less than or equal to x, find CF(6)

40

38

33

5

Answer Details

CF (6) = 5 + 6 + 8 + 10 + 6 + 3
= 38

View Answer

Question 25 Report

In the figure, find the value of x

110o

100o

90o

80o

Answer Details

Z = X = Y = 180o .........(i)

2Z + Y + Y = 190 = 2Z + 2Y = 180

Z + Y = 90o........(ii)

hence x + (z + y) = 180

x + 90o = 180o

x = 180o - 90o

= 90o

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Angles

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Question 26 Report

A train moves from P to Q at an average speed of 90km/h and immediately returns from Q to P through the same route at an average speed of 45km/h. Find the average speed for the entire journey

55.00km/h

60.00km/h

67.50km\h

75.00km\h

Answer Details

Average speed from P to Q = 90km\h
Average speed from O to P = 45km/h
Average for the entire journey = 90 + 45

\frac{135}{2}

= 67.50 km/h

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Question 27 Report

In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40o. Calculate ∠QSR.

30o

40o

45o

50o

Circles

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Question 28 Report

Solve the inequality x - 1 > 4(x + 2)

x > -3

x < -3

2 < x < 3

-3 < x < -2

Answer Details

x - 1 > 4(x + 2) = x - 1 > 4x + 8
4x + 8 < x - 1 = 4x - x < -1 -8

= 3x < -9

∴ x < -3

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Linear Inequalities

Areas

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Question 29 Report

Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

35o

37o

49o

59o

Answer Details

In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)

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Polynomials

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Question 30 Report

The goals scored by 40 football teams from three league divisions are recorded below:

Number of goals	0	1	2	3	4	5	6
frequency	4	3	15	16	1	0	1

What is the total number of goals scored by all the terms?

21

40

91

96

Answer Details

Let x rep. number of goals

\begin{array}{cc} " f & " & f r e q u e n c y \\ x & f & f x \\ 0 & 4 & 0 \\ 1 & 3 & 3 \\ 2 & 15 & 30 \\ 3 & 16 & 48 \\ 4 & 1 & 4 \\ 5 & 0 & 0 \\ 6 & 1 & 6 \end{array}

\sum f x

= 91
Total number of goals scored is

\sum f x

= 91

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Question 31 Report

Find the values of y which satisfy the simultaneous equations x + y = 5, x2 - 2y2 = 1

-12, -2

-12, 12

12, -2

2, -2

Answer Details

x + y = 5.......(i)
x2 - 2y2 = 1.......(ii)
x = 5 - y.........(iii)
Subst. for x in eqn.(ii) = (5 - y)2 - 2y2 = 1
25- 10y + y2 - 2y2 = 1
25 - 1 = y2 + 10y
y2 + 10y2 - 24 = 0
(y + 12)(y - 2) = 0
Then Either y + 12 = 0 or y - 2 = 0
= (12, -2)

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Solution Of Linear Equations

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Question 32 Report

If a u2 - 3v2 and b = 2uv + v2 evaluate (2a - b)(a - b2), when u = 1 and v = -1

9

15

27

33

Answer Details

a = u2 - 3y2 = (1)2 - 3(-1) = -2
b = 2uu + v2 + v2
= 2(1)(-1) + (-1)2 = -1
∴ 2(2a - b)(a - b∴) = [2(-2) - 1](-2 - (-1)2)
= -3 - 3
= 9

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Algebraic Expressions

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Question 33 Report

The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that the edge is 4.3cm

6cm

5cm

4cm

3cm

Answer Details

Base of pyramid of a square of side 8cm vertex directly above the centre edge =

4 \sqrt{3}

cm
From the diagram, the diagonal of one base is AC2 = 82 + 82
Ac2 = 64 + 64 = 128
AC =

8 \sqrt{2}

but OC =

\frac{1}{2}

AC = 8

\sqrt{\frac{2}{2}}

=

4 \sqrt{2}

cm
OE = h = height
h2 = (

4 \sqrt{3}

)2
16 x 2 - 16 x 2
48 - 32 = 16
h =

\sqrt{16}

= 4

View Answer

Question 34 Report

Taking the period of day light on a certain day to be from 5.30a.m to 7.00p.m. Calculate the angle of a pie chart designed to show the periods of the day light and of darkness on that day

187o30' 172o30'

135o225'

202o30' 157o30'

195o165'

Answer Details

Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval
24hrs make 1 day
∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.
Angle of sector of daylight =

\frac{13 \times 60 + 30 \times 360}{(24 \times 60)}

=

\frac{810}{1440}

x 360o
=

\frac{291600}{1440}

= 202.5o
202.5o = 202.30o
darkness period = 360o - 202o 30'
= 157o30'
= 202o30' 157o 30'

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Question 35 Report

The minimum value of $y$ in the equation $y=x^2-6x+8$

8

3

0

-1

5

Quadratic Equations

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Question 36 Report

In the diagram above, QR // TS, QR : TS = 2.3, Find the ratio of the area of triangle PQR to the area of the trapezium QRST.

4:9

4:5

1:3

2:3

Answer Details

h1 =

\frac{h_{1} + h_{2}}{1.5}

=

\frac{h_{1}}{h_{2}} = 2

$\frac{A_{△ P Q R}}{A_{T P (Q R S T)}} = \frac{\frac{1}{2} \times 2 \times h_{1}}{\frac{1}{2} \times h_{2} (2 + 3)}$

$\frac{2}{5} \times \frac{h_{1}}{h_{2}} = \frac{2}{5} \times 2$

= $\frac{4}{5}$

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Mensuration

Triangles And Polygons

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Question 37 Report

Find the values of m which make the following quadratic function a perfect square. x2 + 2(m + 1)x + m + 3

-1, 1

-1, 2

1, -2

2, -2

Answer Details

x2 + 2(m + 1)x + m + 3 from (A + B)2 = a2 + 2AB + b2
2B =

\frac{2 (m + 1)}{2}

= m + 1 .......(i)
Also B2 = m + 3..........(ii)
Subst. for B in equation ..........(iii)
= (m + 1)2 = m + 3 = m2 + 2m + 1 = m + 3
= m2 + 2m - m + 1 - 3 = 0
m2 + m - 2 = 0
(m + 2)(m - 1) = 0 or m - 1 = 0
m = -2 or 1

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Quadratic Equations

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Question 38 Report

Three angles of a nonagon are equal and the sum of six other angles is 1110o. Calculate the size of one of the equal angles.

210o

150o

105o

50o

Answer Details

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Monagon = 9
Where 3 are equal and 6 other angles = 1110o
(2 x 9 - 4)90o = (18 - 4)90o
14 x 90o = 1260o
9 angles = 12600, 6 angles = 110o
Remaining 3 angles = 1260o - 1110o = 150o
Size of one of the 3 angles

\frac{150}{3}

= 50o

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Angles And Intercepts On Parallel Lines

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Question 39 Report

Simplify without using tables $ \frac{2\sqrt{14}\times3\sqrt{21}}{7\sqrt{24}\times2\sqrt{98}} $

3√14

4

3√2

4

3√14

28

3√2

28

82

Answer Details

\frac{2 \sqrt{14} \times 3 \sqrt{21}}{7 \sqrt{24} \times 2 \sqrt{98}}

=

\frac{6 \sqrt{14} \times 3 \times \sqrt{7} \times \sqrt{3}}{7 \times 2 \sqrt{6} \times \sqrt{7} \times \sqrt{14}}

=

\frac{3 \sqrt{3}}{14 \sqrt{6}}

=

\frac{3 \sqrt{3}}{14 \sqrt{2} \times \sqrt{3}}

=

\frac{\sqrt{3}}{\sqrt{2}}

x

\frac{\sqrt{2}}{\sqrt{2}}

=

\frac{3 \sqrt{2}}{28}

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Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Question 40 Report

Given that 1/2 log10 P = 1, find the value Of P

104

103

102

101

10-1

Answer Details

1/2log10P
log10P1/2 = 1
P1/2 = 10
p = 102

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Logarithms

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Question 41 Report

In the figure, PQRS is a rectangle. If the shaded area is 72 sq. cm, find h.

12cm

10cm

8cm

5cm

Answer Details

From the diagram, PQRS is a rectangle

Area of shaded part = 72sq.cm

But 72 = 3h - 4 + 6h - 4 + 4h

= 72 - 16h - 8

= 16h - 72 + 8

=16h = 80

h =

\frac{80}{16}

= 5cm

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Question 42 Report

If the length of a square is increased by 20% while while its width is decreased by 20% to form a rectangle, what is the ratio of the area of the rectangle to the area of the square?

6 : 5

25 : 24

5 : 6

24 : 25

Answer Details

Length and width of a square is 100%
Length increased by 20% and
Width decreased by 20% to form a rectangle
Length of rectangle = 120% to form a rectangle
Length of rectangle = 120% and
Width of rectangle = 80%
Area of rectangle = L x W
Area of square = W
Ratio of the area of the rectangle to the area of the square
A =

\frac{Area of rectangle}{Area of square}

\frac{120 \times 30}{100 \times 100}

=

\frac{96}{100}

= 24 : 25

Read lesson note on Lengths And Perimeters (WAEC)

Lengths And Perimeters

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Question 43 Report

What is the circumference of latitude 0oS if R is the radius of the earth?

R cos

θ

2

π

R cos

θ

R sin

θ

2

π

R sin

θ

Answer Details

Circumference of Latitude 0oS where R is the radius of the earth is = 2

π

R cos

θ

Read lesson note on Circles (WAEC)

Read lesson note on Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

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Question 44 Report

If P varies inversely as V and V varies directly as R2, find the relationship between P and R given that R = 7 when P = 2.

P = 98R2

PR2 = 98

P =

\frac{1}{98 R^{2}}

P =

\frac{P R^{2}}{98}

Answer Details

P =

\frac{1}{v}

and vR2 = P =

\frac{k}{v}

......(i)
and v KR2 .......(ii)
(where k is constant)
Subst. for v in equation (i) = p =

\frac{1^{2}}{K R}

.....(ii)
when r = 7, p = 2
2 =

\frac{k}{7^{2}}

k = 2 x 49
= 98
Subt. foe k in ....(iii)
P =

\frac{98}{R^{2}}

PR2 = 98

Read lesson note on Variation (JAMB)

Read lesson note on Sequence And Series (WAEC)

Variation

Sequence And Series

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Question 45 Report

A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room

5

17

8

17

8

15

12

17

Answer Details

ABCD is the floor, by pythagoras
AC2 = 144 + 81 =

\sqrt{225}

AC = 15cm
Height of room is 8m, diagonal of floor is 15m
∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152
cosine =