Consider a triangle with sides of length 8cm, 12cm and 14cm. What are the sizes of all interior angles of the triangle?

Question

Consider a triangle with sides of length 8cm, 12cm and 14cm.  What are the sizes of all interior angles of the triangle?

Emery Adkins · Accepted Answer

The interior angles of triangle are
# /_A= 34.77^0, /_B=58.81^0 , /_C= 86.42^0 # Sides of the triangle are #a=8 , b=12, c=14# cm each. The semi perimeter of triangle is #s=(a+b+c)/2# or #s= (8+12+14)/2 =17# unit Area of the triangle is #A_t=sqrt(s(s-a)(s-b)(s-c)# or #A_t=sqrt(17(17-8)(17-12)(17-14)# or #A_t=sqrt(17*9*5*3)=sqrt(2295)~~47.91# sq.unit Area of the triangle is #47.91# sq.unit. We know Area ,# A_t= 1/2*a*b*sinC# where sides #a=8 , b=12# and their included angle is #/_C# #:. 47.91=1/2*8*12*sin C or sin C= (47.91*2)/(8*12)# #sin C= 0.998045 :. /_C=sin^-1( 0.998045) or /_C= 86.42^0# Similarly # A_t= 1/2*b*c*sinA# where sides #b=12 , c=14# and their included angle is #/_A# #:. 47.91=1/2*12*14*sin A or sin A= (47.91*2)/(12*14)# #sin A= 0.5703 :. /_A=sin^-1( 0.5703) or /_A= 34.77^0# #:. /_B= 180-(86.42+34.77)=58.81^0 # The interior angles of triangle are # /_A= 34.77^0, /_B=58.81^0 , /_C= 86.42^0 # [Ans]