Introduction to trigonometry 109
(i)^12 ×base×perpendicular height, or
(ii)^12 absinCor^12 acsinBor^12 bcsinA,or
(iii)
√
[s(s−a)(s−b)(s−c)],where
s=
a+b+c
2
11.9 Worked problemson the
solution of triangles and
finding their areas
Problem 27. In a triangleXYZ,∠X= 51 ◦,
∠Y= 67 ◦andYZ= 15 .2cm. 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
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
zy
x15.2cm
67
51
Y
X
Z
Figure 11.25
Area of triangleXYZ=^12 xysinZ
=^12 ( 15. 2 )( 18. 00 )sin 62◦=120.8cm^2 (or area
=^12 xzsinY=^12 ( 15. 2 )( 17. 27 )sin 67◦=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 and
XZis the longest of the three sides.
Problem 28. Solve the trianglePQRand find its
area given thatQR= 36 .5mm,PR=29.6mm and
∠Q= 36 ◦.
TrianglePQRis shown in Fig. 11.26.
P
r
QRp36.5mm
q29.6mm
36
Figure 11.26
Applying the sine rule:
29. 6
sin36◦
=
36. 5
sinP
from which,
sinP=
36 .5sin36◦
29. 6
= 0. 7248
HenceP=sin−^10. 7248 = 46 ◦ 27 ′or 133◦ 33 ′.
WhenP= 46 ◦ 27 ′andQ= 36 ◦then
R= 180 ◦− 46 ◦ 27 ′− 36 ◦= 97 ◦ 33 ′.
WhenP= 133 ◦ 33 ′andQ= 36 ◦then
R= 180 ◦− 133 ◦ 33 ′− 36 ◦= 10 ◦ 27 ′.
Thus, in this problem, there aretwoseparate sets of
results and both are feasible solutions. Such a situation
is called theambiguous case.
Case 1.P= 46 ◦ 27 ′,Q= 36 ◦,R= 97 ◦ 33 ′,
p=36.5mm andq= 29 .6mm.
From the sine rule:
r
sin97◦ 33 ′
=
29. 6
sin36◦
from which,
r=
29 .6sin97◦ 33 ′
sin36◦
=49.92 mm
Area=^12 pqsinR=^12 ( 36. 5 )( 29. 6 )sin97◦ 33 ′
=535.5mm^2
Case 2.P= 133 ◦ 33 ′,Q= 36 ◦,R= 10 ◦ 27 ′,
p=36.5mm andq= 29 .6mm.