Answered By Jonny C

A.) The 5ft ladder is being propper against a straight wall, it makes a triangle.

5 ft = Hypoteneus

1 ft = Adjacent

Theta = ?

Cos (Theta) = 1/5

Theta = Arccos (0.2)

Theta = 78.5 degrees, since this exceeds 75 degrees the ladder is not being used safely.

B.) To find the minimum distance, we use 75 degrees as theta, as that's the cutoff point to which the ladder can be used safely.

Theta = 75 degrees

Hypoteneus = 10 ft

Adjacent = ?

Cos (75) = A/10

10cos(75) = A

A = 2.59 ft, so the minimum distance is 2.59 ft.

Answered By Clifton P

We have a straight ladder, wall, and ground(hopefully not too many bumps) so we have a triangle with at least two sides known. The ladder length, 5ft, and the distance from the wall, 1ft.

A) is looking for what angle is being made by the ladder and if it falls within the safe range given.

We can use one of SOH CAH or TOA to figure it out. Since we have the hypotenuse(ladder), and the ground length which is adjacent to the angle we want to find, CAH is our friend.

CAH ->  $\cos\theta=\frac{Adjacent}{Hypotenuse}=\frac{1ft}{5ft}$cosθ=AdjacentHypotenuse =1ƒ t5ƒ t  

rearrange to solve for  $\theta$θ 

 $\cos^{-1}\left(\cos\theta\right)=\theta=\cos^{-1}\left(\frac{1}{5}\right)\approx78.5^{\circ}$cos⁻¹(cosθ)=θ=cos⁻¹(15 )≈78.5^? 

We're given that the angle needs to be LESS than  $75^{\circ}$75^?  so we can say the ladder is not in the safe range

 

B) The other ladder is 10ft long and this time we have the angle and need to find the adjacent side. CAH is still our go to friend here.

 $\cos\theta=\frac{A}{H}$cosθ=AH  

Using the minimum safety angle of  $75^{\circ}$75^? 

And new hypotenuse of 10ft

 $\cos\left(75\right)=\frac{A}{10ft}$cos(75)=A10ƒ t  

*a little rearranging*

 $10ft\cdot\cos\left(75\right)\approx2.6ft$10ƒ t·cos(75)≈2.6ƒ t 

This is our minimum safe distance for the base of the ladder from the wall.