If sin θ = $\dfrac{3}{5}$ and θ is acute angle, find

(i) cos θ

(ii) tan θ.

(i) By formula,

⇒ sin² θ + cos² θ = 1

Substituting values we get,

$\Rightarrow \Big(\dfrac{3}{5}\Big)^2 + \text{cos}^2 \text{ θ} = 1 \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = 1 - \dfrac{9}{25} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{25 - 9}{25} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{16}{25} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\dfrac{16}{25}} \\[1em] \Rightarrow \text{cos θ} = \pm \dfrac{4}{5}.$

Since, θ is an acute angle and value of cos is positive in first quadrant.

∴ cos θ = $\dfrac{4}{5}$.

Hence, cos θ = $\dfrac{4}{5}$.

(ii) By formula,

tan θ = $\dfrac{\text{sin θ}}{\text{cos θ}}$

= $\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} = \dfrac{3}{4}$.

Hence, tan θ = $\dfrac{3}{4}$.