Higher Engineering Mathematics, Sixth Edition

110 Higher Engineering Mathematics

From the sine rule:

r

sin10◦ 27 ′

=

29. 6

sin36◦

from which,

r=

29 .6sin10◦ 27 ′

sin36◦

=9.134mm

Area=^12 pqsinR=^12 ( 36. 5 )( 29. 6 )sin10◦ 27 ′

=97.98mm^2.

TrianglePQRfor case 2 is shown in Fig. 11.27.

P

QR

36 

133  33 

10  27 

29.6mm

36.5mm

9.134mm

Figure 11.27

Now try the following exercise

Exercise 49 Further problems on solving

triangles and finding their areas

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

trianglesABCand find their areas.

1. A= 29 ◦,B=[ 68 ◦,b=27mm.
   C= 83 ◦,a= 14 .1mm,
   c= 28 .9mm,area=189mm^2

]

2. 
   

   B= 71 ◦ (^26) [′,C= 56 ◦ 32 ′,b= 8 .60cm.
   A= 52 ◦ 2 ′,c= 7 .568cm,
   a= 7 .152cm,area= 25 .65cm^2
   ]
   In Problems 3 and 4, use the sine rule to solve the
   trianglesDEFand find their areas.

   
   


3. 
   

   d=17cm, f=22cm,F= 26 ◦.
   [
   D= 19 ◦ 48 ′,E= 134 ◦ 12 ′,
   e = 36 .0cm,area=134cm^2

   
   

]

4. d= 32 .6mm,e= 25 .4mm,D= 104 ◦ 22 ′.
   [
   E= 49 ◦ 0 ′,F= 26 ◦ 38 ′,
   f= 15 .09mm,area= 185 .6mm^2

]

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 ◦.

⎢

⎢

⎢

⎣

J= 44 ◦ 29 ′,L= 99 ◦ 31 ′,

l= 5 .420cm,area= 6 .132cm^2 or

J= 135 ◦ 31 ′,L= 8 ◦ 29 ′,

l= 0 .811cm, area= 0 .917cm^2

⎤

⎥

⎥

⎥

⎦

6. k=46mm,⎡ l=36mm,L= 35 ◦.

⎢⎢

⎢

⎣

K= 47 ◦ 8 ′,J= 97 ◦ 52 ′,

j= 62 .2mm, area= 820 .2mm^2 or

K= 132 ◦ 52 ′,J= 12 ◦ 8 ′,

j= 13 .19mm, area= 174 .0mm^2

⎤

⎥⎥

⎥

⎦

11.10 Further worked problemson

solving triangles and finding

their areas

Problem 29. Solve triangleDEFand find its area

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

∠E= 64 ◦.

TriangleDEFis shown in Fig. 11.28.

D

E

e

d35.0mm F

f25.0mm

64 

Figure 11.28

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. 1 = 1083

from which,e=

√

1083 =32.91mm

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 ′