Question

In ∆ABC, given angle A=70° and angle C=40. If CD is perpendicular to AB. Then angle ACD= ___________. * 30° 40° 50° 20°

Answer 1

Answer:

In the triangle ABC , 3. Angle BDC = 90°.........given

Therefore Angle CBD +BDC +BCD=180.......Sum of angles of triangle

40°+90°+BCD=180°

130°+BCD=180°

BCD=50°

2. Angle ACB=90°......given

Angle ACB=Angle ACD+BCD

90°=ACD+50°

Angle ACD=40°

1. Angle ABC=40°............(A-D-B)

ACB+ABC+BAC=180°............Sum of angles of triangle

Therefore 90°+40°+BAC=180°

130°+BAC=180°

BAC=50°

mark me as brainlist plz.....

guys plz like and follow me.....

Answer 2

$\huge\bold\red{❤Answer❤}$

50 degree