Question

given ab and bc are tangents of d and the acd=15 what is the measure of abc?

Answer 1

Answer on Question #60109 – Math – Geometry

Question

Given AB and BC are tangents of the circle with center D and the $\angle ACD = 15{}^\circ$, what is the measure of $\angle ABC$?

Solution

Triangle $\triangle ACD$ is isosceles, because $AD$ and $CD$ have the same measure as radii of circle, hence

$$\angle ACD = \angle CAD = 15{}^\circ.$$

Using the triangle sum property, 'the total measure of the interior angles in any triangle is $180{}^\circ$' calculate

$$\angle ADC = 180{}^\circ - 15{}^\circ - 15{}^\circ = 150{}^\circ.$$

Tangents AB and BC always form a right angle with the circle's radius, therefore,

$$\angle DAB = \angle DCB = 90{}^\circ.$$

Using the statement, 'the total measure of the interior angles of a quadrilateral is $360{}^\circ$' finally obtain

$$\angle ABC = 360{}^\circ - \angle ADC - \angle DAB - \angle DCB = 360{}^\circ - 150{}^\circ - 90{}^\circ - 90{}^\circ = 30{}^\circ.$$

Answer: $30{}^\circ$

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