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.
- A= 29 ◦,B=[ 68 ◦,b=27mm.
C= 83 ◦,a= 14 .1mm,
c= 28 .9mm,area=189mm^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.
d=17cm, f=22cm,F= 26 ◦.
[
D= 19 ◦ 48 ′,E= 134 ◦ 12 ′,
e = 36 .0cm,area=134cm^2
]
- 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.
- 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
⎤
⎥
⎥
⎥
⎦
- 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 ′