Two ladders of length a lean against opposite walls of an alley with their feet touching. One ladder extends h feet up the wall and makes a 75° angle with the ground. The other ladder extends k feet up the opposite wall and makes a
45° angle with the ground. Find the width of the alley in terms of a, h, and/or k. Assume the ground is horizontal and perpendicular to both walls.
how would we find the width of the alley?
sqrt (a^2 - k^2) + sqrt (a^2 - h^2) ?
Hi Blank,
Here is the pic. (maybe next time you could display some of your own working for us)
Note that a=a' and k_1=k (because it is an isosceles triangle)
I added g, it is g that you need to find in terms of the original letters.
Width of alley= k+g
\(cos75= \frac{g}{a}\\ g=a*cos75\)
So width of alley = k + a*cos75
At this point I have answered the question but it is not very difficult to go further.
Using pythagoras a=sqrt(2k^2)
And cos75 can be estimated on the calculator (which is sensible since it is a real life situation)
But the exact value of cos 75 can also be found using the expansion of cos(45+30)