In the figure above, AT is a tangent to the circle drawn from point T, touching the circle at point A, and ACB is a diameter of the circle. If \( \angle BCT = 130^{\circ} \), find \( \angle ATC \).
Show Hint
Angles ACT and BCT lie on a straight line and add to 180 degrees, and a tangent is always perpendicular to the radius drawn to the point of contact.
Step 1: Set up coordinates for the figure.
Place the centre C at the origin (0,0) and let the circle have radius r. Since AT is horizontal and tangent at A, the radius CA must be vertical, because a tangent is always perpendicular to the radius at the point of contact. So place \( A = (0, r) \) and, since ACB is a diameter, \( B = (0, -r) \).
Step 2: Place point T using the given angle.
Since AT is horizontal through A, point T has the same height as A, so \( T = (t, r) \) for some positive t. Ray CB points straight down, at $-90^{\circ}$ from the positive x-axis. Turning $130^{\circ}$ from CB towards CT (matching the figure) places ray CT at $-90^{\circ} + 130^{\circ} = 40^{\circ}$ from the positive x-axis.
Step 3: Solve for t using this direction.
A ray from C at $40^{\circ}$ satisfies $\frac{r}{t} = \tan 40^{\circ}$, since T lies on it. So $t = r \cot 40^{\circ}$.
Step 4: Find vectors from T and use the dot product.
Vector $TA = (0-t,\, r-r) = (-t, 0)$, and vector $TC = (0-t,\, 0-r) = (-t, -r)$.
$$\cos(\angle ATC) = \frac{TA \cdot TC}{|TA||TC|} = \frac{t^2}{t\sqrt{t^2+r^2}} = \frac{t}{\sqrt{t^2+r^2}}$$
Step 5: Substitute t and simplify using an identity.
With $t = r\cot 40^{\circ}$, we get $t^2 + r^2 = r^2(\cot^2 40^{\circ} + 1) = r^2 \csc^2 40^{\circ}$, using $\cot^2\theta + 1 = \csc^2\theta$. So $\sqrt{t^2+r^2} = \frac{r}{\sin 40^{\circ}}$.
$$\cos(\angle ATC) = \frac{r\cot 40^{\circ}}{r/\sin 40^{\circ}} = \cot 40^{\circ} \times \sin 40^{\circ} = \cos 40^{\circ}$$
Final Answer:
So $\angle ATC = 40^{\circ}$, the same answer confirmed here using coordinates and vectors instead of pure angle chasing.
\[ \boxed{40^{\circ}} \]
