Basic Engineering Mathematics, Fifth Edition

Non-right-angled triangles and some practical applications 209

Now try the following Practice Exercise

PracticeExercise 91 Solution of triangles

and their areas (answers on page 350)

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 ◦


2. q= 3 .25m,r= 4 .42m,P= 105 ◦
   In problems 3 and 4, use the cosine and sine rules
   to solve the trianglesXYZand find their areas.


3. x= 10 .0cm,y= 8 .0cm,z= 7 .0cm


4. x=21mm,y=34mm,z=42mm

23.5 Practical situations involving trigonometry

There are a number ofpractical situationsin which the
use of trigonometryis needed tofind unknownsides and
anglesoftriangles.Thisisdemonstratedinthefollowing
worked problems.

Problem 7. 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

A section of the roof is shown in Figure 23.9.

B

A 8.0m C

338 408

Figure 23.9

Angle at ridge,B= 180 ◦− 33 ◦− 40 ◦= 107 ◦

From the sine rule,

8. 0

sin107◦

=

a

sin33◦

from which a=

8 .0sin33◦

sin107◦

= 4 .556m=BC

Also from the sine rule,

8. 0

sin107◦

=

c

sin40◦

from which c=
8 .0sin40◦
sin107◦

= 5 .377m=AB

Hence,the roof slopes are 4.56m and 5.38m, correct

to the nearest centimetre.

Problem 8. A man leaves a point walking at

6 .5km/hinadirectionE20◦N (i.e. a bearing of

70 ◦). A cyclist leaves the same point at the same

time in a direction E 40◦S (i.e. a bearing of 130◦)

travelling at a constant speed. Find the average

speed of the cyclist if the walker and cyclist are

80km apart after 5 hours

After5hoursthewalkerhastravelled5× 6. 5 = 32 .5km

(shown asABin Figure 23.10). IfACis the distance the

cyclist travels in 5 hours thenBC=80km.

b

A

B

C

80km

W

S

408 E

N^208

32.5km

Figure 23.10

Applying the sine rule,

80

sin60◦

=

32. 5

sinC

from which sinC=

32 .5sin60◦

80

= 0. 3518

Hence, C=sin−^10. 3518 = 20. 60 ◦

(or 159. 40 ◦, which is not possible)

and B= 180 ◦− 60 ◦− 20. 60 ◦= 99. 40 ◦

Applying the sine rule again,

80

sin60◦

=

b

sin99. 40 ◦

from which b=

80sin99. 40 ◦

sin60◦

= 91 .14km

Since the cyclist travels 91.14km in 5 hours,

average speed=

distance

time

=

91. 14

5

= 18 .23km/h

Problem 9. Two voltage phasors are shown in

Figure 23.11. IfV 1 =40V andV 2 =100V,

determine the value of their resultant (i.e. length

OA) and the angle the resultant makes withV 1