Explanation:
I take it then that all 3 lumbers (the central vertical one, and the 2 left and right slanted ones) have 1 ft in diameter.
the scenario creates 4 large right-angled triangles (left and right inner and outer ones).
and we need a few small ones on the ends of the slanted beams to get the distances there.
let's start on the right side :
inner triangle A
leg1 = 3 ft
leg2 = tbd
Hypotenuse = tbd
angle1 (opposite of leg1) = 30°
angle3 (opposite of Hypotenuse) = 90°
angle2 = 180 - 90 - 30 = 60°
remember the law of sine
a/sin(A) = b/sin(B) = c/sin(C)
with a,b,c being the sides opposite of the corresponding angles A, B, C.
3/sin(30) = Hypotenuse/sin(90) = Hypotenuse
3/0.5 = Hypotenuse
Hypotenuse = 6 ft
Hypotenuse² = leg1² + leg2²
6² = 3² + leg2²
36 = 9 + leg2²
25 = leg2²
leg2 = 5 ft
outer triangle A
the angles are the same as the inner triangle.
the sides increase by the side lengths of small right-angled triangle at the end of the beam.
at both ends of the inner Hypotenuse we draw right-angled heights to the outer edge of the beam.
these heights are 1 ft (remember, 1 ft diameter for each beam).
the bottom right little triangle has also 30°, 60°, 90° angles, the extension of leg2 of the inner triangle A is the Hypotenuse of that small triangle, one leg is the 1 ft diameter of the beam, the other leg is the lower extension of the beam of the outer triangle A.
1/sin(30) = extension of leg2iA = 1/0.5 = 2 ft
leg2oA = 5 + 2 = 7 ft
leg1oA² = 2² - 1² = 3
leg1oA = sqrt(3) = 1.732050808... ft
now for the upper extensions with the upper right little triangle. it had also the angles 30°, 60°, 90°.
1/sin(60) = extension of leg1iA = 1.154700538... ft
leg1oA = 3 + 1.154700538... = 4.154700538... ft
lenghtA² = leg1oA² + leg2oA² =
= 4.154700538...² + 7² =
= 17.26153656... + 49 =
= 66.26153656...
lengthA = 8.140118953... ft
inner triangle B
leg1 = 3 ft
leg2 = 3 ft
(45° angle in right-angled triangle means isoceles triangle)
Hypotenuse² = 3² + 3² = 18
Hypotenuse = sqrt(18) = 4.242640687... ft
since the small triangle at both ends of length B are also isoceles triangles with 1 ft legs
lengthB = Hypotenuse + 1 + 1 = 6.242640687... ft
the extension to the bottom inner leg is
Hypotenuse² = 1² + 1² = 2
Hypotenuse = sqrt(2) = 1.414213562... ft
Dim. X = 15 - 1 - 7 - 1 - 3 - 1.414213562... = 1.585786438... ft
length A ≈ 8.14 ft
length B ≈ 6.24 ft
Dim. X ≈ 1.59 ft