question_answer

A, B, C are three points on a circle. The tangent at A meets BC produced at T, \[\angle BTA=40{}^\circ ,\]\[\angle CAT=44{}^\circ .\]The angle subtended by BC at the centre of the circle is

A) \[84{}^\circ \]

B) \[92{}^\circ \]

C) \[96{}^\circ \]

D) \[104{}^\circ \]

Correct Answer: D

Solution :

[d] In \[\Delta ACT,\] \[\angle ACB\,\,\,\,180{}^\circ -\angle CAT\,\,\,\,\angle ATC\] \[=180{}^\circ -(44{}^\circ +40{}^\circ )\] \[=180{}^\circ -84{}^\circ =96{}^\circ \] \[\therefore \]      \[\angle ACB\,\,180{}^\circ -\angle ACT\] \[=180{}^\circ -96{}^\circ =84{}^\circ \] Also, \[\angle \,ACB\,\,\angle CAT=44{}^\circ \] \[\therefore \] In \[\Delta ABC,\] \[\angle BCA\ 180{}^\circ -(\angle ABC\,\,\angle ACB)\] \[\Rightarrow \]   \[\angle BCA\ 180{}^\circ -(44{}^\circ +84{}^\circ )\] \[=180{}^\circ -128{}^\circ =52{}^\circ \] \[\therefore \]      \[\angle BOC=2\angle BAC=2\times 52{}^\circ =104{}^\circ \]