Higher Engineering Mathematics, Sixth Edition

Introduction to trigonometry 99

(vi) cotangentθ=

adjacent side

opposite side

i.e. cotθ=

a

b



c

b

a

Figure 11.6

(b) From above,

(i)

sinθ

cosθ

=

b

c

a

c

=

b

a

=tanθ,

i.e. tanθ=

sinθ

cosθ

(ii)

cosθ

sinθ

=

a

c

b

c

=

a

b

=cotθ,

i.e. cotθ=

cosθ

sinθ

(iii) secθ=

1

cosθ

(iv) cosecθ=

1

sinθ

(Note ‘s’and‘c’ go together)

(v) cotθ=

1
tanθ
Secants, cosecants and cotangents are called the
reciprocal ratios.

Problem 3. If cosX=

9

41

determine the value of

the other five trigonometry ratios.

Fig. 11.7 shows a right-angled triangleXY Z.

Y

Z

X 9

41

Figure 11.7

Since cosX=

9

41

,thenXY=9 units and

XZ=41 units.

Using Pythagoras’ theorem: 41^2 = 92 +YZ^2 from

whichYZ=

√

( 412 − 92 )=40 units.

Thus

sinX=

40

41

,tanX=

40

9

= 4

4

9

,

cosecX=

41

40

= 1

1

40

,

secX=

41

9

= 4

5

9

andcotX=

9

40

Problem 4. If sinθ= 0 .625 and cosθ= 0. 500

determine, without using trigonometric tables or

calculators, the values of cosecθ,secθ,tanθ

and cotθ.

cosecθ=

1

sinθ

=

1

0. 625

=1.60

secθ=

1

cosθ

=

1

0. 500

=2.00

tanθ=

sinθ

cosθ

=

0. 625

0. 500

=1.25

cotθ=

cosθ

sinθ

=

0. 500

0. 625

=0.80

Problem 5. PointAlies at co-ordinate (2, 3) and

pointBat (8, 7). Determine (a) the distanceAB,

(b) the gradient of the straight lineAB, and (c) the

angleABmakes with the horizontal.

(a) PointsAandBare shown in Fig. 11.8(a).

In Fig. 11.8(b), the horizontal and vertical lines

ACandBCare constructed.

Since ABC is a right-angled triangle, and

AC=( 8 − 2 )=6andBC=( 7 − 3 )=4, then by

Pythagoras’ theorem

AB^2 =AC^2 +BC^2 = 62 + 42

and AB=

√

( 62 + 42 )=

√

52 =7.211,

correct to 3 decimal places.

(b) The gradient ofABisgivenbytanA,

i.e.gradient=tanA=

BC

AC

=

4

6

=

2

3

(c) TheangleABmakeswiththehorizontalisgiven

by tan−^123 =33.69◦.