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Question
Given : sin A = `(3)/(5)` , find : (i) tan A (ii) cos A
Solution
Consider the diagram below : 
sin A = `(3)/(5)`
i .e. `"perpendicular"/"hypotenuse" = (3)/(5) ⇒"BC" /"AC" = (3)/(5)`
Therefore if length of BC = 3x, length of AC = 5x
Since
AB2 + BC2 = AC2 ...[ Using Pythagoras Theorem ]
AB2 + (3x)2 = (5x)2
AB2 = 25x2 – 9x2 = 16x2
∴ AB = 4x (base)
Now
(i) tan A = `"perpendicular"/"base" = (3x)/(4x) = (3)/(4)`
(ii) cos A = `"base"/"hypotenuse" = (4x)/(5x) = (4)/(5)`
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