In a \( \Large \triangle ABC \), \( \Large \angle A \) : \( \Large \angle B \) : \( \Large \angle C \) = 2 : 3 : 4. A line CD drawn parallel to AB, then \( \Large \angle ACD \) is


A) \( \Large 80 ^{\circ} \)

B) \( \Large 20 ^{\circ} \)

C) \( \Large 40 ^{\circ} \)

D) \( \Large 60 ^{\circ} \)

Correct Answer:
C) \( \Large 40 ^{\circ} \)

Description for Correct answer:

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]

\( \Large \therefore \angle ACD = 40 ^{\circ} \)


Part of solved Geometry questions and answers : >> Elementary Mathematics >> Geometry








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