A ladder AB, which is set up with B on the horizontal ground and A against a vertical wall, makes an angle of 71 degrees with the ground. It a prevented from slipping by a tight rope PQ of length 1.68metres, fixed at P to the foot of the wall and at Q to the ladder, PQ being at right angles to AB. Calculate in centimeters, The height |AP| of the ladder above P, The length of the ladder

5151.48 cms

Step-by-step explanation:

We have to:

1.68 m is 1680 cm.

Now the rope forms another right triangle. PQB, we can find the value of QB as follows:

tan (71 °) = PQ / QB

Solving:

QB = PQ / tan (71 °)

Replacing we have:

QB = 1680 / 2.9

QB = 579.31

Knowing this value we can find QA using similar triangles PQB and AQP, like this:

QA / PQ = PQ / QB

Solved:

QA = PQ ^ 2 / QB

Replacing:

QA = 1680 ^ 2 / 579.31

QA = 4872

Finally, knowing QA, we can calculate the one we want to know AP:

Sin (71 °) = AP / AB

But AB = QA + QB = 4872 + 579.31, therefore AB = 5451.31

Solved:

AP = Sin (71 °) * AB

Replacing:

AP = 0.945 * 5451.31

AP = 5151.48 cms

the correct answer is x/3+10≥32

hope this !