110 Higher Engineering Mathematics
From the sine rule:
r
sin 10 ◦ 27 =
29.6
sin 36 ◦
from which,
r =
29.6 sin10 ◦ 27
sin 36 ◦
= 9.134 mm
Area =
1
2 pq sin R =
1
2 (36.5)(29.6) sin 10
◦ 27
= 97.98 mm
2
.
Triangle PQR for case 2 is shown in Fig. 11.27.
P
Q
R
36Њ
133Њ33 Ј
10Њ27 Ј
29.6 mm
36.5 mm
9.134 mm
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
triangles ABC and find their areas.
1. A = 29 ◦ , B = 68 ◦ , b = 27 mm.
C = 83 ◦ , a =14.1 mm,
c = 28.9 mm, area = 189 mm 2
2. B = 71 ◦ 26 , C = 56 ◦ 32 , b = 8.60 cm.
A = 52 ◦ 2 , c = 7.568 cm,
a = 7.152 cm, area = 25.65 cm 2
In Problems 3 and 4, use the sine rule to solve the
triangles DEF and find their areas.
3. d = 17 cm, f = 22 cm, F = 26 ◦ .
D = 19 ◦ 48 , E =134 ◦ 12 ,
e = 36.0 cm, area = 134 cm 2
4. d = 32.6 mm, e = 25.4 mm, D = 104 ◦ 22 .
E = 49 ◦ 0 , F = 26 ◦ 38 ,
f = 15.09 mm, area = 185.6 mm
2
In Problems 5 and 6, use the sine rule to solve the
triangles JKL and find their areas.
5. j = 3.85 cm, k = 3.23 cm, K = 36 ◦ .
⎡
⎢
⎢
⎢
⎣
J = 44 ◦ 29 , L = 99 ◦ 31 ,
l = 5.420 cm, area = 6.132 cm
2 or
J = 135 ◦ 31 , L = 8 ◦ 29 ,
l = 0.811 cm, area = 0.917 cm 2
⎤
⎥
⎥
⎥
⎦
6. k = 46 mm, l = 36 mm, L =35 ◦ .
⎡
⎢
⎢
⎢
⎣
K = 47 ◦ 8 , J = 97 ◦ 52 ,
j = 62.2 mm, area = 820.2 mm 2 or
K = 132 ◦ 52 , J = 12 ◦ 8 ,
j = 13.19 mm, area = 174.0 mm 2
⎤
⎥
⎥
⎥
⎦
11.10 Further worked problems on
solving triangles and finding
their areas
Problem 29. Solve triangle DEF and find its area
given that EF = 35.0 mm, DE = 25.0 mm and
∠E = 64 ◦ .
Triangle DEF is shown in Fig. 11.28.
D
E
e
F
d ϭ 35.0 mm
f ϭ 25.0 mm
64 Њ
Figure 11.28
Applying the cosine rule:
e
2
= d
2
+ f
2
− 2d f cos E
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 = 32.91 mm
Applying the sine rule:
32.91
sin 64 ◦ =
25.0
sin F
from which, sin F =
25.0 sin64 ◦
32.91
= 0.6828
Thus
∠F = sin
−1 0.6828
= 43
◦ 4
or 136
◦ 56
From the sine rule:
r
sin 10 ◦ 27 =
29.6
sin 36 ◦
from which,
r =
29.6 sin10 ◦ 27
sin 36 ◦
= 9.134 mm
Area =
1
2 pq sin R =
1
2 (36.5)(29.6) sin 10
◦ 27
= 97.98 mm
2
.
Triangle PQR for case 2 is shown in Fig. 11.27.
P
Q
R
36Њ
133Њ33 Ј
10Њ27 Ј
29.6 mm
36.5 mm
9.134 mm
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
triangles ABC and find their areas.
1. A = 29 ◦ , B = 68 ◦ , b = 27 mm.
C = 83 ◦ , a =14.1 mm,
c = 28.9 mm, area = 189 mm 2
2. B = 71 ◦ 26 , C = 56 ◦ 32 , b = 8.60 cm.
A = 52 ◦ 2 , c = 7.568 cm,
a = 7.152 cm, area = 25.65 cm 2
In Problems 3 and 4, use the sine rule to solve the
triangles DEF and find their areas.
3. d = 17 cm, f = 22 cm, F = 26 ◦ .
D = 19 ◦ 48 , E =134 ◦ 12 ,
e = 36.0 cm, area = 134 cm 2
4. d = 32.6 mm, e = 25.4 mm, D = 104 ◦ 22 .
E = 49 ◦ 0 , F = 26 ◦ 38 ,
f = 15.09 mm, area = 185.6 mm
2
In Problems 5 and 6, use the sine rule to solve the
triangles JKL and find their areas.
5. j = 3.85 cm, k = 3.23 cm, K = 36 ◦ .
⎡
⎢
⎢
⎢
⎣
J = 44 ◦ 29 , L = 99 ◦ 31 ,
l = 5.420 cm, area = 6.132 cm
2 or
J = 135 ◦ 31 , L = 8 ◦ 29 ,
l = 0.811 cm, area = 0.917 cm 2
⎤
⎥
⎥
⎥
⎦
6. k = 46 mm, l = 36 mm, L =35 ◦ .
⎡
⎢
⎢
⎢
⎣
K = 47 ◦ 8 , J = 97 ◦ 52 ,
j = 62.2 mm, area = 820.2 mm 2 or
K = 132 ◦ 52 , J = 12 ◦ 8 ,
j = 13.19 mm, area = 174.0 mm 2
⎤
⎥
⎥
⎥
⎦
11.10 Further worked problems on
solving triangles and finding
their areas
Problem 29. Solve triangle DEF and find its area
given that EF = 35.0 mm, DE = 25.0 mm and
∠E = 64 ◦ .
Triangle DEF is shown in Fig. 11.28.
D
E
e
F
d ϭ 35.0 mm
f ϭ 25.0 mm
64 Њ
Figure 11.28
Applying the cosine rule:
e
2
= d
2
+ f
2
− 2d f cos E
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 = 32.91 mm
Applying the sine rule:
32.91
sin 64 ◦ =
25.0
sin F
from which, sin F =
25.0 sin64 ◦
32.91
= 0.6828
Thus
∠F = sin
−1 0.6828
= 43
◦ 4
or 136
◦ 56