Question

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 \( \angle ATC \).

• A. \( 50^{\circ} \)
• B. \( 60^{\circ} \)
• C. \( 30^{\circ} \)
• D. \( 40^{\circ} \)

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.

Answer 1

The Correct Option is D

Step 1: Use the straight line formed by the diameter.
Since ACB is a diameter, points A, C and B lie on one straight line, with the centre C between A and B. So angle ACT and angle BCT sit on a straight line and add up to \( 180^{\circ} \).
\[ \angle ACT = 180^{\circ} - \angle BCT = 180^{\circ} - 130^{\circ} = 50^{\circ} \]

Step 2: Use the tangent-radius property.
AT is a tangent to the circle at point A, and CA is a radius drawn to that exact point of contact. A tangent line is always perpendicular to the radius at the point where it touches the circle, so \( \angle CAT = 90^{\circ} \).

Step 3: Use the angle sum of triangle ACT.
The three angles of triangle ACT add up to \( 180^{\circ} \).
\[ \angle CAT + \angle ACT + \angle ATC = 180^{\circ} \]
\[ 90^{\circ} + 50^{\circ} + \angle ATC = 180^{\circ} \]
\[ \angle ATC = 40^{\circ} \]

Step 4: Watch the common mistake.
A student who forgets the tangent is perpendicular to the radius might wrongly take \( \angle ATC \) as equal to \( \angle ACT = 50^{\circ} \) and pick option (a). The right angle at A is what brings the answer down to \( 40^{\circ} \).

Final Answer:
\( \angle ATC = 40^{\circ} \). \[ \boxed{40^{\circ}} \]