A 10 foot ladder is leaning against a wall that is 4 ft away.

Question

How tall is the ladder when it touched the wall?

What is the angle measurement between the bottom of the ladder and floor?

What is the angle measurement between the top of the ladder and the wall?

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

Answer 1

Hi Kayla:

Here's the way to tackle this type of problem.

First, draw a careful diagram on the standard X-Y axes.

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

Then, the base of the ladder is at point P (4. 0) and top of ladder meets the wall at point Q (0, y) - we don't yet know where the ladder meets the wall.

However, we do know that ΔOPQ is right angled. So, we can 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) Here we need ∠OQP = 90 - ∠OPQ

= 90 - 66.42

= 23.58 degrees

Part (d) Now OQ = 5 and we are asked to find OP. Again, let's use good old Pythagoras.

As before, OQ² + OP² = PQ²

5² + OP² = 10²

OP² = 10² - 5²

= 100 - 25

=> OP = √75

= 5√3

= ~8.66 feet