Question

In the figure, circle with centre D touches the sides
of angleACB at A and B as shown. If angleACB=52°
then find the measure of angle ADB ​

Answer 1

Answer:

Given ; In quadrilateral ADBC, AC and BC are two tangents and DA and DB is the two radius.

By question,

AC=BC [tangents]

DA=DB [radius]

angle CAD =90° [radius at the point of contact is

perpendicular to tangent]

angle DBC =90° [radius at the point of contact is

perpendicular to tangent]

therefore,

angle,ADB+DBC+ABC+CAD =360° [sum angles of

quadrilateral

=360°]

angle,ADB+90°+52°+90° = 360°

angle, ADB+232° =360°

angle ,ADB = 360°-232°

therefore, angle ADB = 128°