Basic Engineering Mathematics, Fifth Edition

208 Basic Engineering Mathematics

from which e=

√

1082. 85

= 32 .91mm=DF

Applying the sine rule,

32. 91

sin64◦

=

25. 0

sinF

from which sinF=

25 .0sin64◦

32. 91

= 0. 6828

Thus, ∠F=sin−^10. 6828 = 43 ◦ 4 ′or 136◦ 56 ′

F= 136 ◦ 56 ′is notpossible inthis case since 136◦ 56 ′+

64 ◦is greater than 180◦. Thus, onlyF= 43 ◦ 4 ′is valid.

Then∠D= 180 ◦− 64 ◦− 43 ◦ 4 ′= 72 ◦ 56 ′.

Area of triangleDEF=

1

2

dfsinE

=

1

2

( 35. 0 )( 25. 0 )sin 64◦= 393 .2mm^2

Problem 5. A triangleABChas sides

a= 9 .0cm,b= 7 .5cmandc= 6 .5cm. Determine

its three angles and its area

TriangleABCis shown in Figure 23.7. It is usual first

to calculate the largest angle to determine whether the

triangle is acute or obtuse. In this case the largest angle

isA(i.e. opposite the longest side).

A

B a 5 9.0cm C

c 5 6.5cm b 5 7.5cm

Figure 23.7

Applying the cosine rule, a^2 =b^2 +c^2 − 2 bccosA

from which 2 bccosA=b^2 +c^2 −a^2

and cosA=

b^2 +c^2 −a^2

2 bc

=

7. 52 + 6. 52 − 9. 02

2 ( 7. 5 )( 6. 5 )

= 0. 1795

Hence, A=cos−^10.^1795 =^79.^67 ◦

(or 280. 33 ◦, which is clearly impossible)

The triangle is thus acute angled since cosAis positive.

(If cosAhad been negative, angleAwould be obtuse;

i.e., would lie between 90◦and 180◦.)

Applying the sine rule,

9. 0

sin79. 67 ◦

=

7. 5

sinB

from which sinB=

7 .5sin79. 67 ◦

9. 0

= 0. 8198

Hence, B=sin−^10. 8198 = 55. 07 ◦

and C= 180 ◦− 79. 67 ◦− 55. 07 ◦= 45. 26 ◦

Area=

√

[s(s−a)(s−b)(s−c)], where

s=

a+b+c

2

=

9. 0 + 7. 5 + 6. 5

2

= 11 .5cm

Hence,

area=

√

[11. 5 ( 11. 5 − 9. 0 )( 11. 5 − 7. 5 )( 11. 5 − 6. 5 )]

=

√

[11. 5 ( 2. 5 )( 4. 0 )( 5. 0 )]= 23 .98cm^2

Alternatively,area=

1

2

acsinB

=

1

2

( 9. 0 )( 6. 5 )sin 55. 07 ◦= 23 .98cm^2

Problem 6. Solve triangleXYZ,shownin

Figure 23.8, and find its area given that

Y= 128 ◦,XY= 7 .2cmandYZ= 4 .5cm

1288

x 5 4.5cm

z 5 7.2cm

y

X

Y Z

Figure 23.8

Applying the cosine rule,

y^2 =x^2 +z^2 − 2 xzcosY

= 4. 52 + 7. 22 −[2( 4. 5 )( 7. 2 )cos 128◦]

= 20. 25 + 51. 84 −[− 39 .89]

= 20. 25 + 51. 84 + 39. 89 = 112. 0

y=

√

112. 0 = 10 .58cm=XZ

Applying the sine rule,

10. 58

sin128◦

=

7. 2

sinZ

from which sinZ=

7 .2sin128◦

10. 58

= 0. 5363

Hence, Z=sin−^10. 5363 = 32. 43 ◦

(or 147. 57 ◦which is not possible)

Thus,X= 180 ◦− 128 ◦− 32. 43 ◦= 19. 57 ◦

Area ofXYZ=

1

2

xzsinY=

1

2

( 4. 5 )( 7. 2 )sin 128◦

= 12 .77cm^2