206 Basic Engineering Mathematics
or (b)
1
2
absinCor
1
2
acsinBor
1
2
bcsinA
or (c)
√
[s(s−a)(s−b)(s−c)]wheres=
a+b+c
2
23.3 Worked problems on the solution
of triangles and their areas
Problem 1. In a triangleXYZ,∠X= 51 ◦,
∠Y= 67 ◦andYZ= 15 .2cm. Solve the triangle and
find its area
The triangleXYZis shown in Figure 23.2. Solving the
triangle means finding∠Zand sidesXZandXY.
X
Y x 5 15.2cm Z
z y
518
678
Figure 23.2
Since the angles in a triangle add up to 180◦,
Z= 180 ◦− 51 ◦− 67 ◦= 62 ◦
Applying the sine rule,
15. 2
sin51◦
=
y
sin67◦
=
z
sin62◦
Using
15. 2
sin51◦
=
y
sin67◦
and transposing gives y=
15 .2sin67◦
sin51◦
= 18 .00cm=XZ
Using
15. 2
sin51◦
=
z
sin62◦
and transposing gives z=
15 .2sin62◦
sin51◦
= 17 .27cm=XY
Area of triangleXYZ=
1
2
xysinZ
=
1
2
( 15. 2 )( 18. 00 )sin 62◦= 120 .8cm^2
(or area=
1
2
xzsinY=
1
2
( 15. 2 )( 17. 27 )sin67◦
= 120 .8cm^2 )
It is always worth checking with triangle problems that
the longest side is opposite the largest angle and vice-
versa. In this problem,Yis the largest angle andXZis
the longest of the three sides.
Problem 2. Solve the triangleABCgiven
B= 78 ◦ 51 ′,AC= 22 .31mm andAB= 17 .92 mm.
Also find its area
TriangleABCis shown in Figure 23.3. Solving the
triangle means finding anglesAandCand sideBC.
A
Ba C
b 5 22.31mm
c^5
17.92mm
788519
Figure 23.3
Applying the sine rule,
22. 31
sin78◦ 51 ′
=
17. 92
sinC
from whichsinC=
17 .92sin78◦ 51 ′
22. 31
= 0. 7881
Hence, C=sin−^10. 7881 = 52 ◦ 0 ′or 128◦ 0 ′
SinceB= 78 ◦ 51 ′,Ccannot be 128◦ 0 ′, since 128◦ 0 ′+
78 ◦ 51 ′is greater than 180◦. Thus, onlyC= 52 ◦ 0 ′is
valid.
AngleA= 180 ◦− 78 ◦ 51 ′− 52 ◦ 0 ′= 49 ◦ 9 ′.
Applying the sine rule, a
sin49◦ 9 ′
=
22. 31
sin78◦ 51 ′
from which a=^22 .31sin49
◦ 9 ′
sin78◦ 51 ′
= 17 .20mm
Hence,A= 49 ◦ 9 ′,C= 52 ◦ 0 ′andBC= 17 .20mm.
Area of triangleABC=
1
2
acsinB
=
1
2
( 17. 20 )( 17. 92 )sin78◦ 51 ′
= 151 .2mm^2
Problem 3. Solve the trianglePQRand find its
area given thatQR= 36 .5mm,PR= 29 .6mmand
∠Q= 36 ◦