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Answer 1 · 2015-11-03T17:50:25

#abs(b+a) = sqrt(3)#

Consider the diagram below.
#vec(b)=vec(PQ)# and #abs(vec(PQ))=abs(vecb)=1#

#veca = vec(PQ)=vec(TS)# and #abs(vec(PQ)) = abs(vec(TS)) = abs(veca)=1#

#vec(a+b) = vec(PS)#

#/_STU = /_QPT = 60^@# (the angle between #vec(a)# and #vec(b)#
#rArr /_PTS = 120^@#
#rArr /_RPT =/_RST=30^@#

The bisector of #/_PTS#
divides #triangle PTS# into 2 congruent triangles
#abs(PR) = abs(RS)#

Since #/_PTR=60^@# and #abs(PT)=1#
#rArr abs(PR) = sqrt(3)/2#

and
#abs(PS) = sqrt(3)#

Therefore #abs(vec(a+b)) = sqrt(3)#

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