There are two noncongruent triangles where B - 55°, a = 15, and b = 13. Find the measures of the angles of the triangle with the
greater perimeter. Round to the nearest tenth if necessary.

I tried to use the law of consines

Step-by-step explanation:

b^2 = a^2 + c^2 - 2ac*cos(B)

13^2 = 15^2 + c^2 - 2*15*c*cos(55)

169 = 225 + c^2 - 30c*cos(55)

c^2 - 30c*cos(55) + 225 - 169 = 0

c^2 - 30c*cos(55) + 56 = 0

c = (30*cos(55) +/- sqrt((30*cos(55))^2 - 4*1*56)) / (2*1)

c = (30*cos(55) +/- sqrt(900*cos^2(55) - 224)) / 2

c = 15*cos(55) +/- sqrt(225*cos^2(55) - 56)

To have the greater perimeter we want the larger value of c:

c = 15*cos(55) + sqrt(225*cos^2(55) - 56)

c =~ 12.848965599926043766708967572196

010

23

step-by-step explanation:

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b