Question

3 rays have common vertex. All the angles, which are made by neighbor rays( I think they mean adjacent rays), can be:

Answer 1

Answer:

$b)\ Obtuse$

Step-by-step explanation:

$We\ are\ given\ that:\\There\ are\ three\ rays.\\Lets\ mark\ them\ Ray\ OP, OQ,\ and\ OR\ with\ their\ common\ Vertex\ O.\\\\Now,\\We\ know\ that\ the\ sum\ of\ all\ angles\ around\ a\ point[or\ here\ Vertex\ O]\\ should\ be\ equal\ to\ 360.\\Here,\\\\Lets\ consider\ each\ case\ sepearately.\\If\ all\ the\ angles\ formed: \angle POR, \angle ROQ, \angle POQ,\ are\ acute\ then:\\\angle POR, \angle ROQ, \angle POQ$

$\angle POR+ \angle ROQ+ \angle POQ$

$But\ here\ the\ sum\ of\ the\ angles\ are\ less\ than\ 270,\ but\ the\ sum\ should\\ instead\ be\ 360.\\Hence,\ the\ angles\ formed\ by\ the\ rays\ are\ NOT\ acute.$

$Similarly,\\If\ all\ the\ angles\ formed: \angle POR, \angle ROQ, \angle POQ,\ are\ right\ then:\\\angle POR, \angle ROQ, \angle POQ =90\\When\ we\ add\ them\ again:\\\angle POR+ \angle ROQ+ \angle POQ=90+90+90=270\\But\ again\ the\ sum\ here\ is\ equal\ to\ 270,\ but\ instead\ the\ sum\ had\ to\\ be\ equal\ to\ 360.\\Hence,\\The\ angles\ formed\ are\ NOT\ right.$

$Lets\ consider\ the\ Last\ Case,\\If\ the\ angles\ formed: \angle POR, \angle ROQ, \angle POQ\ are\ obtuse,\\\angle POR, \angle ROQ, \angle POQ>90\\When\ we\ add\ them:\\\angle POR+ \angle ROQ+ \angle POQ>(90+90+90)>270\\Here,\\The\ angles\ formed\ are\ greater\ than\ 270.\ As\ 360>270,\ the\ angles\ ARE\\ obtuse.$