Question
A carpenter is constructing a triangular roof for a storage shed as shown in the figure.
Part A
isosceles triangle = 15 degrees and 15 degrees
length of bas is 45
How high will the peaks of the rise above in the top of the shed?
b/2 = 22.5
tan(15) = h/22.5
h = 22.5(tan15)
Using tangent
Feet
Part B
After the roof a constructed, it will be covered with an asphalt roof material. The carpenter needs to calculate the combined length of the two slope sides. What will be the total length needed of the roof covering?
a = sqr(b^2 +h^2)
then add side isosceles triangle
am I right?
Part A
isosceles triangle = 15 degrees and 15 degrees
length of bas is 45
How high will the peaks of the rise above in the top of the shed?
b/2 = 22.5
tan(15) = h/22.5
h = 22.5(tan15)
Using tangent
Feet
Part B
After the roof a constructed, it will be covered with an asphalt roof material. The carpenter needs to calculate the combined length of the two slope sides. What will be the total length needed of the roof covering?
a = sqr(b^2 +h^2)
then add side isosceles triangle
am I right?
Answers
A. Correct.
B. Cos15 = 22.5/L1.
L1 = 22.5/Cos15 = 23.3.
Length = L1 + L2 = 23.3 + 23.3 = 46.6
B. Cos15 = 22.5/L1.
L1 = 22.5/Cos15 = 23.3.
Length = L1 + L2 = 23.3 + 23.3 = 46.6
the first part is wrong it is 6 because you use 22.5 divided by cos 15
the answer is phil swift
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