Problem:
An acute isosceles triangle, ABC, is inscribed in a circle. Through B and C, tangents to the circle are drawn, meeting at point D. If ∠ABC=∠ACB=2∠D and x is the radian measure of ∠A, then x=
Answer Choices:
A. 73​π
B. 94​π
C. 115​π
D. 136​π
E. 157​π
Solution:
Angles BAC,BCD and CBD all intercept the same circular arc. Therefore ∠BCD=∠CBD=x and ∠D=π−2x. The given condition now becomes 2π−x​=2(π−2x), which has the solution x=73​π.
OR
Let O be the center of the circle. Then ∠COB=2x and, from the sum of the angles of the quadrilateral COBD, we obtain 2x+∠D=π. The conditions of the problem yield x+4∠D=π to be the sum of the angles of △ABC. Solve these two equations in x and ∠D simultaneously to find x=3π/7.
Query. What is x if â–³ABC is an obtuse isosceles triangle?
