Introduction to trigonometry 109
(i)
1
2 × base × perpendicular height, or
(ii)
1
2 ab sin C or
1
2 ac sin B or
1
2 bc sin A, or
(iii)
√
[s(s − a)(s − b)(s − c)], where
s =
a + b + c
2
11.9 Worked problems on the
solution of triangles and
finding their areas
Problem 27. 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 Fig. 11.25. Since
the angles in a triangle add up to 180 ◦ , then
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
z
y
x ϭ15.2 cm
67Њ
51Њ
Y
X
Z
Figure 11.25
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
vice-versa. In this problem, Y is the largest angle and
XZ is the longest of the three sides.
Problem 28. Solve the triangle PQR and find its
area given that QR = 36.5 mm, PR = 29.6 mm and
∠Q = 36 ◦ .
Triangle PQR is shown in Fig. 11.26.
P
r
Q
R
p ϭ 36.5 mm
q ϭ 29.6 mm
36Њ
Figure 11.26
Applying the sine rule:
29.6
sin 36 ◦ =
36.5
sin P
from which,
sin P =
36.5 sin36 ◦
29.6
= 0.7248
Hence P = sin
−1 0.7248 =46 ◦ 27 or 133 ◦ 33 .
When P = 46 ◦ 27 and Q = 36 ◦ then
R = 180 ◦ − 46 ◦ 27 − 36 ◦ = 97 ◦ 33 .
When P = 133 ◦ 33 and Q =36 ◦ then
R = 180 ◦ − 133 ◦ 33 − 36 ◦ = 10 ◦ 27 .
Thus, in this problem, there are two separate sets of
results and both are feasible solutions. Such a situation
is called the ambiguous case.
Case 1. P = 46 ◦ 27 , Q =36 ◦ , R = 97 ◦ 33 ,
p =36.5 mm and q = 29.6 mm.
From the sine rule:
r
sin 97 ◦ 33 =
29.6
sin 36 ◦
from which,
r =
29.6 sin97 ◦ 33
sin 36 ◦
= 49.92 mm
Area =
1
2 pq sin R =
1
2 (36.5)(29.6) sin 97
◦ 33
= 535.5 mm
2
Case 2. P = 133 ◦ 33 , Q = 36 ◦ , R = 10 ◦ 27 ,
p =36.5 mm and q = 29.6 mm.
(i)
1
2 × base × perpendicular height, or
(ii)
1
2 ab sin C or
1
2 ac sin B or
1
2 bc sin A, or
(iii)
√
[s(s − a)(s − b)(s − c)], where
s =
a + b + c
2
11.9 Worked problems on the
solution of triangles and
finding their areas
Problem 27. 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 Fig. 11.25. Since
the angles in a triangle add up to 180 ◦ , then
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
z
y
x ϭ15.2 cm
67Њ
51Њ
Y
X
Z
Figure 11.25
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
vice-versa. In this problem, Y is the largest angle and
XZ is the longest of the three sides.
Problem 28. Solve the triangle PQR and find its
area given that QR = 36.5 mm, PR = 29.6 mm and
∠Q = 36 ◦ .
Triangle PQR is shown in Fig. 11.26.
P
r
Q
R
p ϭ 36.5 mm
q ϭ 29.6 mm
36Њ
Figure 11.26
Applying the sine rule:
29.6
sin 36 ◦ =
36.5
sin P
from which,
sin P =
36.5 sin36 ◦
29.6
= 0.7248
Hence P = sin
−1 0.7248 =46 ◦ 27 or 133 ◦ 33 .
When P = 46 ◦ 27 and Q = 36 ◦ then
R = 180 ◦ − 46 ◦ 27 − 36 ◦ = 97 ◦ 33 .
When P = 133 ◦ 33 and Q =36 ◦ then
R = 180 ◦ − 133 ◦ 33 − 36 ◦ = 10 ◦ 27 .
Thus, in this problem, there are two separate sets of
results and both are feasible solutions. Such a situation
is called the ambiguous case.
Case 1. P = 46 ◦ 27 , Q =36 ◦ , R = 97 ◦ 33 ,
p =36.5 mm and q = 29.6 mm.
From the sine rule:
r
sin 97 ◦ 33 =
29.6
sin 36 ◦
from which,
r =
29.6 sin97 ◦ 33
sin 36 ◦
= 49.92 mm
Area =
1
2 pq sin R =
1
2 (36.5)(29.6) sin 97
◦ 33
= 535.5 mm
2
Case 2. P = 133 ◦ 33 , Q = 36 ◦ , R = 10 ◦ 27 ,
p =36.5 mm and q = 29.6 mm.