206 Basic Engineering Mathematics
or (b)
1
2
ab sin C or
1
2
ac sin B or
1
2
bc sin A
or (c)
[s(s − a)(s − b)(s − c)] where s =
a + b + c
2
23.3 Worked problems on the solution
of triangles and their areas
Problem 1. In a triangle XYZ, ∠X = 51 ◦ ,
∠Y = 67 ◦ and YZ = 15.2 cm. Solve the triangle and
find its area
The triangle XYZ is shown in Figure 23.2. Solving the
triangle means finding ∠Z and sides XZ and XY.
X
Y
Z
x 515.2 cm
y
z
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
sin 51 ◦ =
y
sin 67 ◦
=
z
sin 62 ◦
Using
15.2
sin 51 ◦ =
y
sin 67 ◦
and transposing gives
y =
15.2 sin67 ◦
sin 51 ◦
= 18.00 cm = XZ
Using
15.2
sin 51 ◦ =
z
sin 62 ◦
and transposing gives
z =
15.2 sin62 ◦
sin 51 ◦
= 17.27 cm = XY
Area of triangle XYZ =
1
2
x y sin Z
=
1
2
(15.2)(18.00) sin 62
◦
= 120.8 cm
2
(or area =
1
2
xz sin Y =
1
2
(15.2)(17.27) sin 67
◦
= 120.8 cm
2
)
It is always worth checking with triangle problems that
the longest side is opposite the largest angle and viceversa. In this problem, Y is the largest angle and X Z is
the longest of the three sides.
Problem 2. Solve the triangle ABC given
B = 78 ◦ 51 , AC = 22.31 mm and AB = 17.92 mm.
Also find its area
Triangle ABC is shown in Figure 23.3. Solving the
triangle means finding angles A and C and side BC.
A
B
a
C
b 5 22.31 mm
c
5 17 .9 2 m m
788519
Figure 23.3
Applying the sine rule,
22.31
sin 78 ◦ 51 =
17.92
sin C
from which sin C =
17.92 sin 78 ◦ 51
22.31
= 0.7881
Hence,
C = sin
−1 0.7881 = 52
◦ 0
or 128
◦ 0
Since B = 78 ◦ 51 , C cannot be 128 ◦ 0 , since 128 ◦ 0 +
78 ◦ 51 is greater than 180 ◦ . Thus, only C = 52 ◦ 0 is
valid.
Angle A = 180 ◦ − 78 ◦ 51 − 52 ◦ 0 = 49 ◦ 9 .
Applying the sine rule,
a
sin 49 ◦ 9 =
22.31
sin 78 ◦ 51
from which
a =
22.31 sin 49 ◦ 9
sin 78 ◦ 51 = 17.20 mm
Hence, A = 49
◦ 9
, C = 52
◦ 0
and BC = 17.20 mm.
Area of triangle ABC =
1
2
ac sin B
=
1
2
(17.20)(17.92) sin 78
◦ 51
= 151.2 mm
2
Problem 3. Solve the triangle PQR and find its
area given that QR = 36.5 mm, PR = 29.6 mm and
∠Q = 36 ◦
or (b)
1
2
ab sin C or
1
2
ac sin B or
1
2
bc sin A
or (c)
[s(s − a)(s − b)(s − c)] where s =
a + b + c
2
23.3 Worked problems on the solution
of triangles and their areas
Problem 1. In a triangle XYZ, ∠X = 51 ◦ ,
∠Y = 67 ◦ and YZ = 15.2 cm. Solve the triangle and
find its area
The triangle XYZ is shown in Figure 23.2. Solving the
triangle means finding ∠Z and sides XZ and XY.
X
Y
Z
x 515.2 cm
y
z
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
sin 51 ◦ =
y
sin 67 ◦
=
z
sin 62 ◦
Using
15.2
sin 51 ◦ =
y
sin 67 ◦
and transposing gives
y =
15.2 sin67 ◦
sin 51 ◦
= 18.00 cm = XZ
Using
15.2
sin 51 ◦ =
z
sin 62 ◦
and transposing gives
z =
15.2 sin62 ◦
sin 51 ◦
= 17.27 cm = XY
Area of triangle XYZ =
1
2
x y sin Z
=
1
2
(15.2)(18.00) sin 62
◦
= 120.8 cm
2
(or area =
1
2
xz sin Y =
1
2
(15.2)(17.27) sin 67
◦
= 120.8 cm
2
)
It is always worth checking with triangle problems that
the longest side is opposite the largest angle and viceversa. In this problem, Y is the largest angle and X Z is
the longest of the three sides.
Problem 2. Solve the triangle ABC given
B = 78 ◦ 51 , AC = 22.31 mm and AB = 17.92 mm.
Also find its area
Triangle ABC is shown in Figure 23.3. Solving the
triangle means finding angles A and C and side BC.
A
B
a
C
b 5 22.31 mm
c
5 17 .9 2 m m
788519
Figure 23.3
Applying the sine rule,
22.31
sin 78 ◦ 51 =
17.92
sin C
from which sin C =
17.92 sin 78 ◦ 51
22.31
= 0.7881
Hence,
C = sin
−1 0.7881 = 52
◦ 0
or 128
◦ 0
Since B = 78 ◦ 51 , C cannot be 128 ◦ 0 , since 128 ◦ 0 +
78 ◦ 51 is greater than 180 ◦ . Thus, only C = 52 ◦ 0 is
valid.
Angle A = 180 ◦ − 78 ◦ 51 − 52 ◦ 0 = 49 ◦ 9 .
Applying the sine rule,
a
sin 49 ◦ 9 =
22.31
sin 78 ◦ 51
from which
a =
22.31 sin 49 ◦ 9
sin 78 ◦ 51 = 17.20 mm
Hence, A = 49
◦ 9
, C = 52
◦ 0
and BC = 17.20 mm.
Area of triangle ABC =
1
2
ac sin B
=
1
2
(17.20)(17.92) sin 78
◦ 51
= 151.2 mm
2
Problem 3. Solve the triangle PQR and find its
area given that QR = 36.5 mm, PR = 29.6 mm and
∠Q = 36 ◦