Question 1196263
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This is one way to draw out the diagram
*[illustration diagram]
Let x be the length of segment AD
And let y be the height of the hut (i.e. the length of segment CD)


Triangle BCD is a 45-45-90 triangle
The two legs (BD and CD) are the same length
So BD = y and CD = y


AB = 15
AD = x
BD = AB - AD
BD = 15 - x


Since BD = y was mentioned earlier, and BD = 15-x, we can say y = 15-x


For triangle ACD, we can say:
tan(angle) = opposite/adjacent
tan(A) = CD/AD
tan(55) = y/x
x*tan(55) = y
y = x*tan(55)


Now plug in y = 15-x and solve for x
y = x*tan(55)
15-x = x*tan(55)
15 = x*tan(55)+x
x*tan(55)+x = 15
x*(tan(55)+1) = 15
x = 15/(tan(55)+1)
x = 6.17754764468651


Then use this to find the value of y
y = 15-x
y = 15-6.17754764468651
y = 8.8224523553135
y = 8.822452
Or,
y = x*tan(55)
y = 6.17754764468651*tan(55)
y = 8.82245235531349
y = 8.822452


Answer: Approximately 8.822452 meters


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