make a sketch, let the shed touch the wall at A and the ground at B
Let the ladder touch the wall at P and the ground at Q
Let R be the point where the ladder touches the shed
let O be the bottom of the wall meeting the ground.
I am also going to scale down the values given by a factor of 50 to keep the numbers smaller.
let's use trig: let angle Q = angle P = Ø
then cosØ = 2/PR ----> PR = 2secØ
and sinØ = 5/RQ ---> RQ = 5cscØ
PQ = PR+RQ
= 2secØ + 5cscØ
d(PQ)/dØ = 2secØtanØ - 5cscØcotØ
= 0 for a min of PQ
2secØtanØ = 5cscØcotØ
2(1/cosØ)(sinØ/cosØ) = 5(1/sinØ)(cosØ/sinØ)
2sinØ/cos^2 Ø = 5cosØ/sin^2 Ø
sin^3 Ø/cos^3 Ø = 5/2
tan^3 Ø = 2.5
tanØ = 1.3572...
Ø = 53.627°
PQ = ......
let me know what you got, don't forget to multiply PQ by 50, my scaling factor.
A tool shed 250cm high and 100 cm deep is built against a wall. Calculate the shortest ladder that can reach from the ground, over the shed, to the wall behind.
How can I solve this?
1 answer
1 answer