Higher Engineering Mathematics

INTRODUCTION TO TRIGONOMETRY 125

B

a^2 =b^2 +c^2 − 2 bccosA

or b^2 =a^2 +c^2 − 2 accosB

or c^2 =a^2 +b^2 − 2 abcosC

The rule may be used only when:

(i) 2 sides and the included angle are initially given,
or
(ii) 3 sides are initially given.

12.8 Area of any triangle

Thearea of any trianglesuch asABCof Fig. 12.19
is given by:

(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

12.9 Worked problems on the solution

of triangles and finding their

areas

Problem 21. In a triangle XYZ,∠X= 51 ◦,

∠Y= 67 ◦andYZ= 15 .2 cm. Solve the triangle

and find its area.

The triangleXYZ is shown in Fig. 12.20. 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 sin 67◦

sin 51◦

=18.00 cm=XZ

Using

15. 2

sin 51◦

=

z

sin 62◦

and transposing gives:

z=

15 .2 sin 62◦

sin 51◦

=17.27 cm=XY

Figure 12.20

Area of triangleXYZ=^12 xysinZ

=^12 (15.2)(18.00) sin 62◦=120.8 cm^2 (or area

=^12 xzsinY=^12 (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,Yis the largest angle and

XZis the longest of the three sides.

Problem 22. Solve the triangle PQR and

find its area given thatQR= 36 .5 mm,PR=

29.6 mm and∠Q= 36 ◦.

TrianglePQRis shown in Fig. 12.21.

Figure 12.21

Applying the sine rule:

29. 6

sin 36◦

=

36. 5

sinP

from which,

sinP=

36 .5 sin 36◦

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.