Question

A.B and C are three point on circle with centre O. The tangent at C meet BA produced at T. Given that $\angle ATC={36}^{\circ }and\angle ACT={48}^{\circ }$. Calculate the angle subtended by AB at the centre of the circle ?

Answer 1

$\angle ACT={48}^o,\angle ATC={36}^o$

$\therefore \angle CAT={180}^o-({48}^o+{36}^o)$

$={180}^o-{84}^o={96}^o$

$\therefore CAB={180}^o-{96}^o={84}^o\therefore CAB={180}^o-{96}^o={84}^o$

$Now,\angle ABC=\angle ACT={48}^o$

$\therefore \angle BCA={180}^o-(\angle ABC+\angle CAB)\therefore \angle BCA={180}^o-(\angle ABC+\angle CAB)$

$=1{80}^o-({48}^o+{84}^o)={48}^o$

$\therefore \angle BOA=2\times \angle BCA=2\times {48}^o={96}^o$