Higher Engineering Mathematics

134 GEOMETRY AND TRIGONOMETRY

Figure 13.3

From Pythagoras’ theorem,r=

√

42 + 32 =5.

By trigonometric ratios,α=tan−^134 = 36. 87 ◦or

0 .644 rad.

Henceθ= 180 ◦− 36. 87 ◦= 143. 13 ◦or

θ=π− 0. 644 = 2 .498 rad.

Hence the position of pointPin polar co-ordinate

form is (5, 143.13◦) or (5, 2.498 rad).

Problem 3. Express (−5,−12) in polar

co-ordinates.

A sketch showing the position (−5,−12) is shown
in Fig. 13.4.

r=

√

52 + 122 = 13

and α=tan−^1

12

5

= 67. 38 ◦or 1.176 rad

Hence θ= 180 ◦+ 67. 38 ◦= 247. 38 ◦or

θ=π+ 1. 176 = 4 .318 rad

Figure 13.4

Thus (−5,−12) in Cartesian co-ordinates corres-
ponds to (13, 247.38◦) or (13, 4.318 rad) in polar
co-ordinates.

Problem 4. Express (2,−5) in polar

co-ordinates.

A sketch showing the position (2,−5) is shown in

Fig. 13.5.

r=

√

22 + 52 =

√

29 = 5 .385 correct to

3 decimal places

α=tan−^1

5

2

= 68. 20 ◦or 1.190 rad

Henceθ= 360 ◦− 68. 20 ◦= 291. 80 ◦or

θ= 2 π− 1. 190 = 5 .093 rad

Figure 13.5

Thus (2,−5) in Cartesian co-ordinates corres-

ponds to (5.385, 291.80◦) or (5.385, 5.093 rad) in

polar co-ordinates.

Now try the following exercise.

Exercise 61 Further problems on changing

from Cartesian into polar co-ordinates

In Problems 1 to 8, express the given Carte-

sian co-ordinates as polar co-ordinates, correct

to 2 decimal places, in both degrees and in

radians.

1. (3, 5) [(5.83, 59.04◦) or (5.83, 1.03 rad)]


2. (6.18, 2.35)

[

(6.61, 20. 82 ◦)or

(6.61, 0.36 rad)

]

3. (−2, 4)

[

(4.47, 116. 57 ◦)or

(4.47, 2.03 rad)

]