GeminiHeat wrote:
Attachment:
4322.jpg
In the figure above, a circle with center O is tangent to AB at point D and tangent to AC at point C. If \(m∠A = 40°\), then x=
(A) 140
(B) 145
(C) 150
(D) 155
(E) 160
Sum of interior angles in \(Quad. ACOD = 360°\)
Given:
\(∠A = 40°\)
\(∠ACO = ∠ADO = 90°\) ............. [Tangents]
\(40° + 90° + 90° + x = 360°\)
\(x = 140°\)
Hence, option A