Question

in the given figure, triangle ABCD is a right angled triangle at C and CD perpendicular AB if B is 40 degree,find Angle BAC,Angle ACD Angle BCD​

Answer 1

angle D = 90°

Angle B = 40°

Angle A = 90+40+x = 180° (in ∆ CBD)

130+x = 180°

x = 180-130 = 50° = Angle C

Angle C = 50°

Angle D = 90°

Angle A = 50+90+X = 180°

140+X = 180

X = 180-140= 40 = A