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For a certain hyperbola (x^2/a^2)-(y^2/b^2)=1
where a is greater than b, the angle between the asymptotes is  60 degrees. Find a/b

 Mar 20, 2019
 #1
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tan (theta / 2)   =   tan (60/ 2)  =  tan (30)  =   1 /sqrt (3)  =  b / a

 

So

 

tan (60)  =  sqrt (3) / 1  =   a  / b

 

 

 

cool cool cool

 Mar 20, 2019

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