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AB and BC are tangents of the circle with the center at D. The measure of ACD = 15°. Which is the measure of ABC?
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Anyone
can you post a diagram
ok this is what you should know tangents from a external point are equal.. so AB = BC and tangents and radii intersect at right angles... so angle BAD = BCD ( both 90) triangle CDA is isosceles ( since CD and CD are equal radii) so if ACD = 15 then Angle CAD = 15 ( bass angles of an isosceles triangle) now you can find angle ADC from angle sum of a triangle and then you can find angle ABC by angle sum of a quadrilateral ABC = 360 - 90 - 90 - ADC hope it helps
Wait so it's 165?
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