The vertex angle of an isosceles triangle is 40°20'. Each of the leg measures 320 mm. Find the length of the base.
2 answers
each half of the base is 320 sin 20°10'
A = B = (180-40.33)/2 = 69.83o = Base angles.
sinA/a = sinC/c
sin69.83/320 = sin40.33/c
c*sin69.83 = 320*sin40.33
c = 320*sin40.33/sin69.83 = 220.6 mm
Or
cosA = 0.5*AB/AC
AC*cosA = 0.5*AB
AB = 2AC*cosA = 2*320*cos669.83=220.7 mm
sinA/a = sinC/c
sin69.83/320 = sin40.33/c
c*sin69.83 = 320*sin40.33
c = 320*sin40.33/sin69.83 = 220.6 mm
Or
cosA = 0.5*AB/AC
AC*cosA = 0.5*AB
AB = 2AC*cosA = 2*320*cos669.83=220.7 mm
3 answers