Question

In the following figure, circle with centre D touches the sides of \(\angle\)ACB at A and B. If \(\angle\)ACB = 52\(^\circ\), find measure of \(\angle\)ADB. 

Hint:

For a circle touched by two tangents from an external point, the angle between the tangents and the angle between the radii to the points of contact are always supplementary (add up to 180\(^\circ\)). So, you can quickly find \(\angle ADB\) by calculating \(180^\circ - 52^\circ = 128^\circ\).

Answer 1

Step 1: Understanding the Concept: 
This problem involves the properties of tangents to a circle from an external point. The radius to the point of tangency is perpendicular to the tangent line. This forms a quadrilateral whose angle sum property can be used. 
 

Step 2: Key Formula or Approach: 
1. Tangent-Radius Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, \(\angle DAC = 90^\circ\) and \(\angle DBC = 90^\circ\). 

2. Sum of angles in a quadrilateral: The sum of the interior angles of a quadrilateral is 360\(^\circ\). 
 

Step 3: Detailed Explanation: 
Given: \[\begin{array}{rl} \bullet & \text{A circle with centre D touches the sides CA and CB of \(\angle\)ACB at points A and B respectively.} \\ \bullet & \text{CA and CB are tangents to the circle.} \\ \bullet & \text{DA and DB are radii of the circle.} \\ \bullet & \text{\(m\angle ACB = 52^\circ\).} \\ \end{array}\] According to the tangent-radius theorem: \[ DA \perp CA \implies m\angle DAC = 90^\circ \] \[ DB \perp CB \implies m\angle DBC = 90^\circ \] Now, consider the quadrilateral DACB. The sum of its interior angles is 360\(^\circ\). \[ m\angle DAC + m\angle ACB + m\angle CBD + m\angle ADB = 360^\circ \] Substitute the known values: \[ 90^\circ + 52^\circ + 90^\circ + m\angle ADB = 360^\circ \] \[ 232^\circ + m\angle ADB = 360^\circ \] \[ m\angle ADB = 360^\circ - 232^\circ \] \[ m\angle ADB = 128^\circ \]

Step 4: Final Answer: 
The measure of \(\angle\)ADB is 128\(^\circ\).