question_answer

In Fig. 2, AB is the diameter of a circle with centre O and AT is a tangent. If \[\angle AOQ=58{}^\circ \], find \[\angle ATQ\].

Answer:

Given, AB is a diameter of a circle with centre O and AT is a. tangent, then

\[BA\bot AT\]

Also                  \[\angle ABQ=\frac{1}{2}\angle APQ\]

(\[\because \] Angle subtended on the arc is half of the angle subtended at centre)

\[\Rightarrow \angle ABQ=\frac{1}{2}58{}^\circ =29{}^\circ \]

Now,                 \[\angle ATQ=180{}^\circ -(\angle ABQ+\angle BAT)\]

\[=180{}^\circ -(29{}^\circ +90{}^\circ )\]

\[\therefore \angle ATQ=61{}^\circ \]