NEED CORRECTION

A ladder 15 m long leans against the wall. The ladder slides down the wall at a rate of 1.5 m/s. What is the rate of change of the angle at ground when the ladder is 6 m from the wall?

15 y X Using SOHCAHTOA cos 0 = 15 Differentiating both sides with respect to 't' d X dt dt . 15 a cos 0 = - dx at cose = 6/15 dt . 15 - sin 0 - - 1 dx dt 15 . dt de dx dt 15sin 0 . dt Using Pythagoras theorem x+ y = 15' The ladder is 6m from the wall so x = 6m Therefore 6' + y = 15 y = V225 - 36 = V189 And sin 0 = sin 0 = V189 15 Hence de 1.5 dt 15. V189 15 de -1.5 -0.25 Therefore at = - 0. 109 rad/s V189