If sqrt 15 sinθ = 3, find the value of tanθ .

\(\sqrt {15}\) sinθ = 3

sinθ = \(\frac{3}{\sqrt {15}}\)  (where 3 → P and \(\sqrt {15}\) → H)

B = \(\sqrt {(\sqrt {15})^2 - (3)^2}\)

B = \(\sqrt {6}\)

Put all the values →

tanθ = \(\frac{P}{B}\) = \(\frac{3}{\sqrt {6}}\)