Question

please solve this guys....

Answer 1

ABC is right angled at B
AC = hypotenuse
AC^2 = AB^2 + BC^2 = (8root3)^2 + 8^2 = 64*3 + 64 = 64*(3+1) = 64*4
AC = root (64*4) = 8*2 = 16 cm
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To find angle ACB
Let 'X' be the angle.
Cos X = Adjacent side/Hypot = BC/AC
Cos X = 8/16 = 1/2
X = 60 degree
Angle ACB = 60 degree
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Angle BAC
<ABC + BAC + <ACB = 180 [Angle sum property of triangle]
90 + <BAC + 60 = 180
<BAC = 180-150 = 30 degree
Hence,
<ACB = 60
<BAC = 30
Hope  it helps.