Question 12 Marks
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 $\angle ADB$.
Answer
View full question & answer→In quadrilateral DACB, $\angle DAC = 90^{\circ}$ and $\angle DBC = 90^{\circ}$ (Tangent-Radius theorem).
The sum of angles in a quadrilateral is $360^{\circ}$.
$\angle ADB + \angle DAC + \angle ACB + \angle DBC = 360^{\circ}$
$\angle ADB + 90^{\circ} + 52^{\circ} + 90^{\circ} = 360^{\circ}$
$\angle ADB + 232^{\circ} = 360^{\circ}$
$\angle ADB = 360^{\circ} - 232^{\circ} = 128^{\circ}$
The sum of angles in a quadrilateral is $360^{\circ}$.
$\angle ADB + \angle DAC + \angle ACB + \angle DBC = 360^{\circ}$
$\angle ADB + 90^{\circ} + 52^{\circ} + 90^{\circ} = 360^{\circ}$
$\angle ADB + 232^{\circ} = 360^{\circ}$
$\angle ADB = 360^{\circ} - 232^{\circ} = 128^{\circ}$
