find the value of tan 60 geometrically
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trigonometric ratios of 60°
inn right angled triangle ADB,we have
base=BD=a,perpendicular=AD=√3a,hypotenuse=AB=2a and ∠ABD=60°
tan60°=AD/BD=√3a/a=√3
inn right angled triangle ADB,we have
base=BD=a,perpendicular=AD=√3a,hypotenuse=AB=2a and ∠ABD=60°
tan60°=AD/BD=√3a/a=√3
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trigonometric ratios of 60°
inn right angled triangle ADB,we have
base=BD=a,perpendicular=AD=√3a,hypotenuse=AB=2a and ∠ABD=60°
tan60°=AD/BD=√3a/a=√3
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