How do you solve #sqrt3 tanalpha+1=0# if #alpha# is obtuse?

1 Answer
Binayaka C.
Jun 23, 2018

Solution: #alpha= (5pi)/6 or alpha =150^0#

Explanation:

#sqrt 3 tan alpha =-1 : tan alpha = -1/sqrt 3 #

#tan (-pi/6) = -1/sqrt 3 , tan(pi-pi/6)= -1/sqrt3 #

#:. tan((5pi)/6)= -1/sqrt3 #. Since # alpha# is obtuse angle

#alpha= ((5pi)/6) or alpha =150^0#

Solution: #alpha= (5pi)/6 or alpha =150^0# [Ans]

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