if Sin 0 = 3/5 Then COS 20 = ?
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ΔABC be a right angled at B
Let ∠ACB=θ
Given that, sin θ = 3/5
AB/AC = 3/5
Let AB = 3x
then AC = 5x
In right angled ΔABC,
By Pythagoras theorem,
We get
(5x) ² =(3x) ²+BC²
BC ² =(5x)² −(3x)²
BC ². =(2x)²
BC. =4x
(i) cos θ = Base/ Hypotenuse
= BC / AC
= 4x /5x
= 4/5
(ii) tan θ = perpendicular/Base
= AB/BC
= 3x/4x
=3/4
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