
Mark M. answered 10/22/21
Mathematics Teacher - NCLB Highly Qualified
tan (60º + a) = [tan 60º + tan a] / [1 - tan 60º tan a]
Can you substitute and answer?
Tomas C.
asked 10/22/21Mark M. answered 10/22/21
Mathematics Teacher - NCLB Highly Qualified
tan (60º + a) = [tan 60º + tan a] / [1 - tan 60º tan a]
Can you substitute and answer?
Yefim S. answered 10/22/21
Math Tutor with Experience
tan(60° + a) = (tan60° + tana)/(1 - √tan60°tana) = 2√3/(1 - √3·√3) = 2√3/(- 2) = - √3.
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Jeff K.
tan(60° + a) = (tan60° + tan a)/(1 - tan60°tan a) = (√3/2 +√3) / (1 - √3/2 x √3) = 3√3/2 / (1 - 3/2) = 3√3/2 x -2 = -3√310/24/21