ABSTRACT
Problem 2. Solve the triangle ABC given B =78◦51′, AC =22.31 mm and AB=17.92 mm. Find also its area
Triangle ABC is shown in Fig. 25.3. Applying the sine rule:
22.31 sin 78◦51′
= 17.92 sin C
from which, sin C = 17.92 sin78 ◦51′
22.31 = 0.7881
Hence C 0.7881=52◦0′ or ◦ (see Chapters 23). 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 = 22.31sin49 ◦9′
sin 78◦51′ = 17.20 mm
Hence A=49◦9′, C =52◦0′ and BC =17.20 mm. Area of triangle ABC = 12 ac sin B
= 12 (17.20)(17.92)sin78◦51′ = 151.2 mm2
Problem 3. Solve the triangle PQR and find its area given that QR=36.5 mm, PR=26.6 mm and ∠Q =36◦
Triangle PQR is shown in Fig. 25.4.
