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Two boards, AC and BC, are placed such that they meet at point C. Board AC is 6m long and board BC is 5m long. The angle between the two boards at point C is 30 degrees. Find the distance between the

Mathematics

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Here's how to solve the problem:

Step 1: Find the distance between the bottom edges of the boards (side AB).
We can use the Cosine Rule for triangle ABC. The Cosine Rule states c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C.
Here, c=ABc = AB, a=BC=5 ma = BC = 5\,\text{m}, b=AC=6 mb = AC = 6\,\text{m}, and ∠C=30∘\angle C = 30^\circ.

AB2=AC2+BC2−2(AC)(BC)cos⁡(∠C)AB^2 = AC^2 + BC^2 - 2(AC)(BC) \cos(\angle C) AB2=62+52−2(6)(5)cos⁡(30∘)AB^2 = 6^2 + 5^2 - 2(6)(5) \cos(30^\circ) AB2=36+25−60(32)AB^2 = 36 + 25 - 60 \left(\frac{\sqrt{3}}{2}\right) AB2=61−303AB^2 = 61 - 30\sqrt{3} AB2≈61−30(1.73205)AB^2 \approx 61 - 30(1.73205) AB2≈61−51.9615AB^2 \approx 61 - 51.9615 AB2≈9.0385AB^2 \approx 9.0385 AB=9.0385AB = \sqrt{9.0385} AB≈3.006409 mAB \approx 3.006409\,\text{m}

Correct to three significant figures, the distance AB is:

Part a· 1 step

3.01 m\boxed{3.01\,\text{m}}

  1. Find the angle the longer board makes with the floor (angle θ\theta).

    The longer board is AC (6m), and the angle it makes with the floor is ∠A\angle A (labeled θ\theta in the diagram). We can use the Sine Rule: asin⁡A=csin⁡C\frac{a}{\sin A} = \frac{c}{\sin C}.
    Here, a=BC=5 ma = BC = 5\,\text{m}, c=AB≈3.006409 mc = AB \approx 3.006409\,\text{m}, and ∠C=30∘\angle C = 30^\circ.

    BCsin⁡A=ABsin⁡C\frac{BC}{\sin A} = \frac{AB}{\sin C} 5sin⁡θ=3.006409sin⁡30∘\frac{5}{\sin \theta} = \frac{3.006409}{\sin 30^\circ} sin⁡θ=5sin⁡30∘3.006409\sin \theta = \frac{5 \sin 30^\circ}{3.006409} sin⁡θ=5×0.53.006409\sin \theta = \frac{5 \times 0.5}{3.006409} sin⁡θ=2.53.006409\sin \theta = \frac{2.5}{3.006409} sin⁡θ≈0.83155\sin \theta \approx 0.83155 θ=arcsin⁡(0.83155)\theta = \arcsin(0.83155) θ≈56.25∘\theta \approx 56.25^\circ

    Rounding to one decimal place, the angle is:

Part b· 1 step

56.3∘\boxed{56.3^\circ}

  1. Find the shortest distance between the top edges of the boards and the floor.

    This is the height (hh) from point C to the base AB. We can find the area of the triangle using two different formulas and equate them.
    Area of △ABC=12AC⋅BCsin⁡C\triangle ABC = \frac{1}{2} AC \cdot BC \sin C

    Area=12(6)(5)sin⁡30∘Area = \frac{1}{2} (6)(5) \sin 30^\circ Area=12(30)(0.5)Area = \frac{1}{2} (30) (0.5) Area=7.5 m2Area = 7.5\,\text{m}^2

    Also, Area of △ABC=12×base×height=12AB⋅h\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} AB \cdot h.

    7.5=12(3.006409)h7.5 = \frac{1}{2} (3.006409) h h=2×7.53.006409h = \frac{2 \times 7.5}{3.006409} h=153.006409h = \frac{15}{3.006409} h≈4.9893 mh \approx 4.9893\,\text{m}

    Correct to two significant figures, the shortest distance is:

Part c

5.0 m\boxed{5.0\,\text{m}}

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