Question

NO LINKS!! Please help me with this problem. Part 13​

Answer 1

Answer:

7.65 ft apart (to nearest hundredth)

Step-by-step explanation:

The A frame creates an isosceles triangle with equal legs of 10 ft and a vertex of 45°.  To find how far apart the footings should be, we need to find the base of the triangle.

To find the base, use the cosine rule:

$c^2=a^2+b^2-2ab \cos(C)$

where:

• C is the angle
• a and b are the sides adjacent to the angle C
• c is the side opposite the angle C

So for this triangle:

• a = 10
• b = 10
• c = base
• ∠C = 45°

Substitute these values into the formula and solve for c:

$c^2=10^2+10^2-2(10)(10) \cos(45)$

$\implies c^2=200-200\cos(45)$

$\implies c=\sqrt{200-200\cos(45)}$

$\implies c=7.653668647...$

Therefore, the footings should be 7.65 ft apart (to nearest hundredth)

Answer 2

Answer:

  7.65 ft

Step-by-step explanation:

The distance can be figured using the Law of Cosines, or it can be figured by considering half of the isosceles A-frame to be a right triangle. We choose the latter.

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An altitude of the A-frame bisects the vertex angle so that the angle of interest in our right triangle is 22.5°. Then half the distance between the footings will be given by ...

  Sin = Opposite/Hypotenuse

  Opposite = Hypotenuse × Sin

  half-distance = (10 ft)·sin(22.5°)

  footing spacing = 2 × half-distance = (20 ft)sin(22.5°) ≈ 7.654 ft

The footings for each A-frame should be about 7.65 feet apart.