C-09-Q063mediumsingle_mcqIf tanA=43\tan A = \dfrac{4}{3}tanA=34, sinA=?\sin A = ?sinA=?a35\dfrac{3}{5}53b45\dfrac{4}{5}5ব্যাখ্যাHere opposite =4=4=4, adjacent =3=3=3, and hypotenuse =5=5=5, so sinA=oppositehypotenuse=45\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{4}{5}sinA.