If sin θ =3/5 and θ is acute angle, find
(i) cos θ
(ii) tan θ.
Answer :
Let ∆ ABC be a right angled at B
Let ∠ACB = θ
sin θ = 3/5
AB/AC = 3/5
Let AB = 3x
then AC = 5x
In right angled ∆ ABC,
(5x)2 = (3x)2 + BC2
⇒ BC2 = (5x)2 – (3x)2
⇒ BC2 = (2x)2
⇒ BC = 4x
(i) cos θ = Base/ Hypotenuse
= BC / AC
= 4x /5x
= 4/5
(ii) tan θ = perpendicular/Base
= AB/BC
= 3x/4x
= ¾
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