*** timed ***given that a=16 degrees , b=25 degrees and c=28, solve triangle abc. round to the nearest hundredth. a) c= 139 degrees , b= 18.04 , a= 11.76 b) c= 41 degrees , b= 19.93 , a= 30.94 c) c= 139 degrees , b= 17.76 , a= 18.01 d) c= 41 degrees , b= 11.76 , a= 18.04

Step-by-step explanation:

Given are two angles and one side for triangle ABC. We have to solve for the triangle for unknown sides and angles to the nearest hundredth

Angle C =180-(A+B) = 180-(25+16) = 139

a = c·sin(A)/sin(C) =

=28sin16/sin139 = 11.76395

b = c·sin(B)/sin(C) = 18.03696

Round off to nearest 100th we get

a = 11.76 and b = 18.04 and angle C = 139 degrees.

Here there is only one solution and the solution is option A

because they could drop even lower..