A) \( \Large 80 ^{\circ} \) |
B) \( \Large 20 ^{\circ} \) |
C) \( \Large 40 ^{\circ} \) |
D) \( \Large 60 ^{\circ} \) |
C) \( \Large 40 ^{\circ} \) |
Let the angles be 2x, 3x and 4x.
Then. \( \Large 2x + 3x + 4x = 180 ^{\circ} \)
\( \Large 9x = 180 ^{\circ} => x = 20 ^{\circ} \)
\( \Large \therefore \angle A = 2x = 2 \times 20 = 40 ^{\circ} \)
\( \Large \angle B = 3x = 3 \times 20 = 60 ^{\circ} \)
\( \Large \angle C = 4x = 4 \times 20 = 80 ^{\circ} \)
Now, \( \Large AB \parallel CD\ and\ AC\
be\ the\ transversal. \)
Then, \( \Large \angle BAC = \angle ACD \)
[alternate interior angles]
\( \Large \therefore \angle ACD = 40 ^{\circ} \)