Question

In the following figure O is the center of
circle and $\angle BAC=n^{\circ },\angle OCB=m^{\circ }$, then

• A. $m^{\circ }+n^{\circ }={90}^{\circ }$
• B. $m^{\circ }+n^{\circ }={180}^{\circ }$
• C. $m^{\circ }+n^{\circ }={120}^{\circ }$
• D. $m^{\circ }+n^{\circ }={150}^{\circ }$

Answer 1

The correct option is A $m^{\circ }+n^{\circ }={90}^{\circ }$

Since $\angle BAC=n^{\circ }$
$\Rightarrow \angle BOC=2n^{\circ }$(Angle subtended by a chord at the center is twice of the angle subtended by the chord at any other point)
and $\angle OCB=m^{\circ }$

$=\angle OBC=m^o$ ($OB=OC$)
$\Rightarrow m+m+2n={180}^o$
$\Rightarrow m+n={90}^{\circ }$