112 please show how you got the answer and circle answer Sol

1.12
please show how you got the answer and circle answer.

Solution

Triangle ABD : Angle A = 60 deg ; BD = 26\' ; AD = 20\'

Angle B can be found by applying sine rule:

AD/sinB = BD/sin 60

SinB = AD*sin60/BD = 20*sin60/26 =0.67 ----> Angle B = 41.77 deg

AngleD = 180 -60 -41.77 = 78.23 deg

Applying sine rule we find AB :

sinD/AB =sin60/BD ----> AB = sinD*BD/sin60 = sin78.23*26/sin60 = 29.40\'

In triangle BDC Angle D = AngleA +AngleB = 60 + 41.77 = 101.77 deg ( sum of two interior angles = ext. angle)

Applying cosine rule : BC^2 = BD^2 + CD^2 -2BD*CD*cosD

= 26^2 + 30^2 + 2*26*30cos101.77 = 1894.21

BC = 43.52 \'

Angle C can be found by applying sine rule in triangle BCD

sinC/BD = sinD/BC

sinC = BD*sinD/BC = 26*sin101.77/43.52

Angle C = 35.79 deg