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

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation:

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )