**Given that tan θ = 5/12 and θ is an acute angle, find sin θ and cos θ.**

**Answer :**

Consider ∆ ABC be right angled at B and ∠ACB = θ

tan θ = 5/12

AB/BC = 5/12

Consider AB = 5x and BC = 12x

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (5x)^{2} + (12x)^{2}

AC^{2} = 25x^{2} + 144x^{2} = 169x^{2}

AC^{2} = (13x)^{2}

⇒ AC = 13x

In right angled ∆ ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = 5x/13x = 5/13

In right angled ∆ ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC

cos θ = 12x/13x

= 12/13

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