High School

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A circle centered at O has two radii, OA and OB, and two chords, AD and BD. Secants BC and AC are drawn that intersect each other outside the circle at C. Central angle BOA measures 250 degrees.

In this figure:
- \( m∠BDA = 125° \)
- \( m∠BCA = \) [tex]\boxed{}[/tex]

Answer :

The measure of the angle BCA, m∠BCA = 70°

The steps used to find the measure of the angle BCA are as follows;

The possible circle in the question, obtained from a similar question found through search created using MS Word is attached

Circle theorems states that the angle at the center of a circle is twice the angle at the circumference

The measure of the angle BOA = 250°

The measure of the angle at the circumference, BEA, m∠BEA = 125°

The measure of the interior angle, BOA of the quadrilateral BOAE is 360° - 250° = 110°

The radius of a circle is perpendicular to the tangent of the circle at the point of tangency, thus, m∠OBC = 90° and m∠OAC = 90°

The angle BCA in the quadrilateral OBCA is therefore; m∠BCA = 360° - 90° - 90° - 110°

360° - 90° - 90° - 110° = 70°

m∠BCA = 70°

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Rewritten by : Jeany