In a circle, O is the centre and AOB is the diameter. AT is a tangent to the circle. Line TB intersects the circle at Q. Given that ∠AOQ = 94°, find ∠ATQ.

Correct Answer: 43°

Step-by-Step Solution:

1. Understand the given setup:

Let $O$ be the centre of the circle, and $AOB$ be its diameter.

$AT$ is a tangent line to the circle at point $A$.

Line $TB$ intersects the circle at point $Q$.

We are given that the central angle $\angle AOQ=94°$.

2. Find the angle between the tangent and the radius:

Since $AT$ is a tangent at point $A$ and $AB$ is a diameter (and thus $OA$ is a radius), the angle formed between the tangent line $AT$ and the diameter $AB$ is a right angle:

$\angle TAB=90°$

3. Find $\angle ABQ$ using the Inscribed Angle Theorem:

The central angle subtended by arc $AQ$ is $\angle AOQ=94°$.

The inscribed angle subtended by the same arc $AQ$ at the circumference (at vertex $B$) is $\angle ABQ$.

By the Inscribed Angle Theorem, an inscribed angle is half of the central angle subtending the same arc:

$\angle ABQ=\frac{1}{2}\times \angle AOQ$

$\angle ABQ=\frac{1}{2}\times 94°=47°$

4. Calculate $\angle ATQ$ in right-angled triangle $\Delta TAB$:

Consider the right-angled triangle $\Delta TAB$, where $\angle TAB=90°$.

The sum of angles in $\Delta TAB$ is equal to $180°$:

$\angle ATB+\angle TAB+\angle ABT=180°$

Since line $TB$ passes through point $Q$, $\angle ATQ=\angle ATB$ and $\angle ABT=\angle ABQ=47°$:

$\angle ATQ+90°+47°=180°$

$\angle ATQ=180°-137°=43°$

Thus, $\angle ATQ=43°$.