Question

In $△ABC,$ if $\angle A={90}^{\circ }$ then find $\angle B+\angle C$

• A. ${90}^{\circ }$
• B. ${20}^{\circ }$
• C. ${60}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

$A!={90}^o$

By angle sum property, $A!+B!+C!={180}^o$

$B!+C!={180}^o-A!$

$\therefore B!+C!={180}^o-{90}^o$

$\therefore B!+C!={90}^o$