How far from the wall surface must the ladder be moved so that the top of the ladder touches the wall surface 5ft above the floor?

Hi Kayla:

Here"s the method to tackle this kind of problem.

You are watching: A 10 foot ladder is leaning against a wall

First, attract a careful diagram top top the conventional X-Y axes.

Let the basic of the wall be at the origin, O.

Then, the base of the ladder is at suggest P (4. 0) and top the ladder meets the wall surface at allude Q (0, y) - we don"t yet know where the ladder meets the wall.

However, we do understand that ΔOPQ is right angled. So, we have the right to use Pythagoras to find OQ, the distance up the wall where the ladder meets the wall.

Part (a): OQ2 + OP2 = PQ2

OQ2 + 42 = 102

OQ2 = 102 - 42

OQ2 = 100 - 16

∴ OQ = √84

= 2√21

= ~9.17 feet

Part (b) here we need ∠OPQ.

cos ∠OPQ = OP/PQ

= 4/10

=> ∠OPQ = arc cos (0.4)

= 66.42 degrees

Part (c) right here we need ∠OQP = 90 - ∠OPQ

= 90 - 66.42

= 23.58 degrees

Part (d) currently OQ = 5 and also we room asked to discover OP. Again, let"s use an excellent old Pythagoras.

as before, OQ2 + OP2 = PQ2

52 + OP2 = 102

OP2 = 102 - 52

= 100 - 25

=> OP = √75

= 5√3

= ~8.66 feet

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