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# Trigonometry question

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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) ?

Sep 12, 2021

#1
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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) Sep 13, 2021
edited by Melody  Sep 13, 2021
#2
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Your answer is correct too also it is probably obvious but I accidentally replaced h with f  (I didn't use it anyway)

Sep 13, 2021