Higher Engineering Mathematics, Sixth Edition

Introduction to trigonometry 111

F= 136 ◦ 56 ′is not possible in this case since
136 ◦ 56 ′+ 64 ◦is greater than 180◦. Thus only
F= 43 ◦ 4 ′is valid

∠D= 180 ◦− 64 ◦− 43 ◦ 4 ′= 72 ◦ 56 ′

Area of triangleDEF=^12 dfsinE

=^12 (35.0)(25.0) sin64◦=393.2mm^2.

Problem 30. A triangleABChas sidesa=

9.0cm,b= 7 .5cmandc= 6 .5cm. Determine its

three angles and its area.

TriangleABCis shown in Fig. 11.29. 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).

Applying the cosine rule:

a^2 =b^2 +c^2 − 2 bccosA

from which, 2bccosA=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

A

BCa 5 9.0 cm

c 5 6.5 cm b 5 7.5 cm

Figure 11.29

HenceA=cos−^10. 1795 = 79 ◦ 40 ′(or 280◦ 20 ′,whichis
obviouslyimpossible). The triangle is thus acute angled
since cosAis positive.(If cosAhad been negative, angle
Awould be obtuse, i.e. lie between 90◦and 180◦).
Applying the sine rule:

9. 0

sin79◦ 40 ′

=

7. 5

sinB

from which,

sinB=

7 .5sin79◦ 40 ′

9. 0

= 0. 8198

Hence B=sin−^10. 8198 = 55 ◦ 4 ′

and C=^180 ◦−^79 ◦^40 ′−^55 ◦^4 ′=^45 ◦^16 ′

Area=

√

[s(s−a)(s−b)(s−c)],

where s=a+b+c
2

=

9. 0 + 7. 5 + 6. 5

2

= 11 .5cm

Hencearea

=

√

[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=^12 absinC

=^12 ( 9. 0 )( 7. 5 )sin 45◦ 16 ′=23.98cm^2.

Now try the following exercise

Exercise 50 Further problemson solving

triangles and finding their areas

In Problems 1 and 2, use the cosine and sine

rules to solve the trianglesPQRand find their

areas.

1. q=12cm,r=16cm,P= 54 ◦.
   [
   p= 13 .2cm,Q= 47 ◦ 21 ′,
   R= 78 ◦ 39 ′,area= 77 .7cm^2

]

2. q= 3 .25m,r= 4 .42m,P= 105 ◦.
   [
   p= 6 .127m,Q= 30 ◦ 50 ′,
   R= 44 ◦ 10 ′,area= 6 .938m^2

]

In problems 3 and 4, use the cosine and sine

rules to solve the trianglesXY Zand find their

areas.

3. x= 10 .0cm,y= 8 .0cm,z= 7 .0cm.
   [
   X= 83 ◦ 20 ′,Y= 52 ◦ 37 ′,
   Z= 44 ◦ 3 ′,area= 27 .8cm^2

]

4. x=21mm,y=34mm,z=42mm.
   [
   X= 29 ◦ 46 ′,Y= 53 ◦ 30 ′,
   Z= 96 ◦ 44 ′,area=355mm^2

]

11.11 Practical situations involving

trigonometry

There are a number ofpractical situationswhere the

use of trigonometryis needed tofind unknownsides and

anglesoftriangles.Thisisdemonstratedinthefollowing

problems.

Problem 31. A room 8.0m wide has a span

roof which slopes at 33◦on one side and 40◦on the

other. Find the length of the roof slopes, correct to

the nearest centimetre.