Given that a=43 ,b=82 , and c=28, solve triangle abc. round to the nearest tenth.

C= 55°, b= 33.8 units length , a= 23.3 units length

Step-by-step explanation:

The question is incomplete. The complete data is: a=43°, b=82° and c=28 units length

We know that the addition of the angles of a triangle is 180°, then: c = 180° - 43° - 82° = 55°

From law of sines:

sin(a)/a = sin(c)/c

sin(a)*c/sin(c) = a

sin(43°)*28/sin(55) = a

a = 23.3  units length

sin(b)/b = sin(c)/c

sin(b)*c/sin(c) = b

sin(82°)*28/sin(55) = b

b = 33.8  units length

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐