In a triangle ABC, AB=AC and BC is produced to D such that \angle ACD=x, then what is \angle BAC equal to?
- A. 2x-90^\circ
- B. 2x-180^\circ ✓
- C. 180^\circ-2x
- D. \frac{x}{2}
Correct Answer: B. 2x-180^\circ
Explanation
Since AB=AC, we have \angle ABC = \angle ACB. Angles on a straight line give \angle ACB = 180^\circ - x. Thus, \angle ABC = 180^\circ - x. The sum of angles in \Delta ABC is 180^\circ, so \angle BAC = 180^\circ - (180^\circ - x + 180^\circ - x) = 2x - 180^\circ.
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