Question

in the given figure, AB is a 6m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD is 2.54m, find the length of the ladder. Take √3=1.732

Answer 1

Now let AB be the pole and CD be the ladder inclined at angle 60degree
BD=6-2.54=3.46m
now in triangle DBC
sin60=BD/CD
√3/2=3.46/CD
CD=6.92√3÷3
CD= 6.92×1.732÷3
CD=11.98÷3
CD=4m(approx)
therefore the length of CD is 4m