Question
If sin 28Ā° = 15/17 , then tan 62Ā° = ?
Solution
sin 28Ā° = 15/17Ā = perpendicular/hypotenuse So perpendicular = 15 &Ā hypotenuse = 17 so base = 8Ā But for angle 62Ā° which is complimentary of 28Ā°, perpendicular = 8 & base = 15. So tan 62Ā° = perpendicular/baseĀ = 8/15 Alternate Method sin 28Ā° = 15/17 orĀ cos (90Ā°-28Ā°) = 15/17 or cos 62Ā° = 15/17Ā -----(1) Squaring on both sides, we will get cos2 62Ā° = 225/289 or (1 - sin2 62Ā°) = 225/289 or sin2 62Ā° = 1 - 225/289 = 64/289 = 
So sin 62Ā° =Ā 8/17 ----(2) Hence tan 62Ā° = sin62Ā°/cos62Ā° = = Ā Ā
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