Question

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

Answer 1

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]