If $tan A =\frac{2}{3}$, then find sinA.

$\frac{1}{3}$ $\frac{2}{\sqrt{13}}$ $\frac{2}{3}$ $\frac{3}{\sqrt{13}}$

$\frac{2}{\sqrt{13}}$

tan A = \(\frac{2}{3}\) { we know, tanθ = \(\frac{P}{B}\) } By using pythagoras theorem, P² + B² = H² 2² + 3² = H² H = √13 Now, sinA = \(\frac{P}{H}\) = \(\frac{2}{ √13 }\)