\[
\begin{aligned} & {\left[\begin{array}{l}\text { Angle between the tangent and the chord }\end{array}\right]} \\ & \text { [Inscribed angle is half of the central angle]}\end{aligned}
\]

Secant $A B C$ intercepts arc $B C$, which contains $112^{\circ}$; tangent $A D$ at point of tangency $D$ divides this arc in the ratio $7: 9$. Find $\angle B A D$.
Consider the exterior angle of triangle $A D C$, adjacent to angle $A D C$.

## Solution

Let $K$ be a point lying on the extension of the tangent $A D$ beyond point $D$. The angle $K D C$ is an exterior angle for triangle DAC. Therefore,

$\angle A=\angle K D C-\angle D C B=1 / 2-C D-1 / 2-D B=1 / 2\left(9 / 16-{ }^{7} / 16\right) \cdot 112^{\circ}=7^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0ef0b80dacb6d98eff5ag-05.jpg?height=438&width=878&top_left_y=1782&top_left_x=590)

## Answer

$7^{\circ}$.