Basic Engineering Mathematics, Fifth Edition

Non-right-angled triangles and some practical applications 207

TrianglePQRis shown in Figure 23.4.

q 5 29.6mm

p 5 36.5mm

368

r

QR

P

Figure 23.4

Applying the sine rule,
29. 6
sin36◦

=

36. 5

sinP

from whichsinP=

36 .5sin36◦

29. 6

= 0. 7248

Hence, P=sin−^10. 7248 = 46. 45 ◦or 133. 55 ◦

WhenP= 46. 45 ◦andQ= 36 ◦then
R= 180 ◦− 46. 45 ◦− 36 ◦= 97. 55 ◦
WhenP= 133. 55 ◦andQ= 36 ◦then
R= 180 ◦− 133. 55 ◦− 36 ◦= 10. 45 ◦
Thus, in this problem, there aretwoseparate sets of
results and both are feasible solutions. Such a situation
is called theambiguous case.

Case 1.P= 46. 45 ◦,Q= 36 ◦,R= 97. 55 ◦,
p= 36 .5mmandq= 29 .6mm

From the sine rule,

r

sin97. 55 ◦

=

29. 6

sin36◦

from which r=

29 .6sin97. 55 ◦

sin36◦

= 49 .92mm=PQ

Area ofPQR=

1

2

pqsinR=

1

2

( 36. 5 )( 29. 6 )sin 97. 55 ◦

= 535 .5mm^2

Case 2.P= 133. 55 ◦,Q= 36 ◦,R= 10. 45 ◦,
p= 36 .5mmandq= 29 .6mm

From the sine rule,

r

sin10. 45 ◦

=

29. 6

sin36◦

from which r=
29 .6sin10. 45 ◦
sin36◦

= 9 .134mm=PQ

Area ofPQR=

1

2

pqsinR=

1

2

( 36. 5 )( 29. 6 )sin 10. 45 ◦

= 97 .98mm^2

The trianglePQRfor case 2 is shown in Figure 23.5.

368 10.45 8

133.55 8

9.134 mm 29.6 mm

Q 36.5 mm

P

R

Figure 23.5

Now try the following Practice Exercise

PracticeExercise 90 Solution of triangles

and their areas (answers on page 350)

In problems 1 and 2, use the sine rule to solve the

trianglesABCand find their areas.

1. A= 29 ◦,B= 68 ◦,b=27mm


2. B= 71 ◦ 26 ′,C= 56 ◦ 32 ′,b= 8 .60cm
   In problems 3 and 4, use the sine rule to solve the
   trianglesDEFand find their areas.


3. d=17cm,f=22cm,F= 26 ◦


4. d= 32 .6mm,e= 25 .4mm,D= 104 ◦ 22 ′
   In problems 5 and 6, use the sine rule to solve the
   trianglesJKLand find their areas.


5. j= 3 .85cm,k= 3 .23cm,K= 36 ◦


6. k=46mm,l=36mm,L= 35 ◦

23.4 Further worked problems on the

solution of triangles and their

areas

Problem 4. Solve triangleDEFand find its area

given thatEF= 35 .0mm,DE= 25 .0mmand

∠E= 64 ◦

TriangleDEFis shown in Figure 23.6. Solvingthe trian-

gle means finding anglesDandFand sideDF.Since

two sides and the angle in between the two sides are

given, the cosine needs to be used.

648

D

E F

e

d 5 35.0 mm

f 5 25.0 mm

Figure 23.6

Applying the cosine rule, e^2 =d^2 +f^2 − 2 dfcosE

i.e. e^2 =( 35. 0 )^2 +( 25. 0 )^2 −[2( 35. 0 )( 25. 0 )cos 64◦]

= 1225 + 625 − 767. 15

= 1082. 85