As shown in figure, triangle C A B CAB is formed by three tangents to the circle with centre ′ O ′ ′ O ′ . If ∠ A C B

In the given problem, triangle $\triangle CAB$ is formed by three tangents to a circle with center $O$. We know that:

1. $\angle ACB = 50^\circ$, which is an angle between two tangents from points $A$ and $B$ on the circle.
2. Since $A$ and $B$ are points of tangency, $OA = OB$, making $\triangle OAB$ an isosceles triangle.

To find $\angle AOB$, we can use the fact that the angle $\angle ACB$ at the external triangle formed by tangents is half the angle at the center, $\angle AOB$, due to the tangent-segment theorem in a circle.

Solution Steps:

1. According to the tangent-segment theorem, the angle $\angle ACB$ at the circumference (formed by tangents) is half of $\angle AOB$, which is subtended by the same arc $AB$.

   $\angle ACB = \frac{1}{2} \angle AOB$

2. Substituting the given value $\angle ACB = 50^\circ$:

   $50^\circ = \frac{1}{2} \angle AOB$

3. Solving for $\angle AOB$:

   $\angle AOB = 2 \times 50^\circ = 100^\circ$

Answer:

$\angle AOB = 100^\circ$

Would you like further details or explanations on this solution?

Here are some related questions to deepen your understanding:

1. What would $\angle AOB$ be if $\angle ACB$ were $60^\circ$ instead?
2. How does the tangent-segment theorem apply to different configurations of tangent lines?
3. What is the relationship between an angle at the circumference and its corresponding angle at the center in general?
4. Can we apply a similar approach if the tangents form a quadrilateral around the circle?
5. How does the length of the tangent segments from a single external point relate to each other?

Tip: For circles, angles subtended by the same arc are congruent at any point on the circle's circumference.