Question

5. Once Tony reached the hospital, he is carried on a stretcher through the corridors to the emergency area. Two corridors meet at right angles and are 2 m and 3 m wide respectively. 𝜃 is the angle marked on the given figure. AB is the stretcher which must be kept horizontal and cannot be bent as it moves around the corner from one corridor to the other. The stretcher is 4 m long. Determine the greatest length of a stretcher that is able to be horizontally carried around the corner, and thus determine if the 4-m long stretcher is able to get through or not.

Answer 1

i.imgur.com/HvnYEsA.png
here's a photo of the question for reference

Answer 2

Since we can't see where in your diagram you placed 𝜃, on mine at placed 𝜃 at A
of stretcher AB and A touching the wall of the 3m wide corridor.
So I now have two right-angled triangles both containing 𝜃.

I will use trig.

AB = 3sec𝜃 + 2csc𝜃
d(AB)/d𝜃 = 3sec𝜃tan𝜃 - 2csc𝜃cot𝜃 = 0 for a max of AB
3(1/cos𝜃)(sin𝜃/cos𝜃) - 2(1/sin𝜃)(cos𝜃/sin𝜃) = 0
3 sin𝜃/cos^2 𝜃 = 2cos𝜃/sin^2 𝜃
sin^3 𝜃 / cos^3 𝜃 = 2/3
tan^3 𝜃 = 2/3
tan𝜃 = (2/3)^(1/3) = .87358...
𝜃 = 41.139..°

sub back into AB = ....
I got AB = appr 7.02 m

so yes, it will make it

check my arithmetic

Answer 3

Ahhh, your diagram is the same as mine, so all is good