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. π 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.
Ahhh, your diagram is the same as mine, so all is good
i.imgur.com/HvnYEsA.png
here's a photo of the question for reference
Since we can't see where in your diagram you placed π, on mine at placed π 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 π.
I will use trig.
AB = 3secπ + 2cscπ
d(AB)/dπ = 3secπtanπ - 2cscπcotπ = 0 for a max of AB
3(1/cosπ)(sinπ/cosπ) - 2(1/sinπ)(cosπ/sinπ) = 0
3 sinπ/cos^2 π = 2cosπ/sin^2 π
sin^3 π / cos^3 π = 2/3
tan^3 π = 2/3
tanπ = (2/3)^(1/3) = .87358...
π = 41.139..Β°
sub back into AB = ....
I got AB = appr 7.02 m
so yes, it will make it
check my arithmetic
To determine the greatest length of a stretcher that can be horizontally carried around the corner, we need to use trigonometry. Let's break down the problem step by step:
Step 1: Understanding the problem
We have two corridors meeting at right angles. One corridor is 2 m wide, and the other is 3 m wide. We are given a stretcher labeled AB, which must be kept horizontal and cannot be bent as it moves around the corner. The stretcher is 4 m long. We need to determine if the 4-m long stretcher can get through the corner and, if not, find the greatest length of a stretcher that can.
Step 2: Analyzing the corner
Since the stretcher must be kept horizontal and cannot be bent, we can imagine tracing the path of the stretcher as it moves from one corridor to the other. This path will form a right-angled triangle.
Step 3: Identifying the sides of the triangle
In our triangle, the corridors' widths will form the two sides of the triangle. Let's label them as follows:
Side A: 2 m (width of one corridor)
Side B: 3 m (width of the other corridor)
Step 4: Finding the hypotenuse
The hypotenuse of the triangle represents the greatest length of the stretcher that can be carried horizontally around the corner. Let's label it as follows:
Side C: Hypotenuse (the length of the stretcher)
Step 5: Applying trigonometry
We can use the trigonometric relationship of sine to solve this problem. The sine of an angle (π) in a right-angled triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.
sin(π) = Opposite/Hypotenuse
In our triangle, the opposite side of angle π is the width of one corridor (Side A = 2 m), and the hypotenuse is the length of the stretcher (Side C = ? m).
Step 6: Solving for the hypotenuse
Rearranging the sine formula, we can solve for the hypotenuse:
Hypotenuse = Opposite / sin(π)
Side C = Side A / sin(π)
Side C = 2 m / sin(π)
Step 7: Determining if the 4-m stretcher can fit
To determine if the 4-m stretcher can fit around the corner, we need to check if the hypotenuse (Side C) is greater than or equal to 4 m.
If Side C is greater than or equal to 4 m, then the stretcher can fit.
If Side C is less than 4 m, then the stretcher cannot fit.
Step 8: Calculating the greatest length of the stretcher
To calculate the greatest length of the stretcher that can fit, we need to find the maximum value of Side C.
Using the formula: Side C = 2 m / sin(π)
We can vary the angle π and calculate the corresponding values of Side C. The maximum value of Side C will represent the greatest length of the stretcher that can fit around the corner.
Step 9: Conclusion
Compare the maximum value of Side C obtained in Step 8 with the length of the stretcher. If the maximum Side C is greater than or equal to 4 m, then the 4-m stretcher can fit around the corner. Otherwise, it cannot fit.