Whatever the triangle, the sum of its 3 angles is always 180 °.

In the triangle ABC, you can write that:

30 + 105 + α = 180

α = 45 °

In the triangle OBC, you can write that:

β + α + 90 = 180

β = 90 - α → recall: α = 45 °

β = 45 °

In the triangle OAB, you can write that:

90 + 30 + γ = 180

γ = 60 °

In the triangle ABC, you can write that:

AB.cos(30) + BC.cos(45) = 10 → recall: cos(30) = (√3)/2

AB.[(√3)/2] + BC.cos(45) = 10 → recall: cos(45) = (√2)/2

AB.[(√3)/2] + BC.[(√2)/2] = 10

(AB.√3 + BC.√2)/2 = 10

AB.√3 + BC.√2 = 20

BC.√2 = 20 - AB.√3 ← memorize this result

In the triangle ABC, you can write that:

AB.cos(γ) = BC.cos(β) → recall: γ = 60 °

AB.cos(60) = BC.cos(β) → recall: β = 45 °

AB.cos(60) = BC.cos(45) → recall: cos(60) = 1/2

AB.(1/2) = BC.cos(45) → recall: cos(45) = (√2)/2

AB.(1/2) = BC.[(√2)/2]

AB = BC.√2 → recall the memorized result: BC.√2 = 20 - AB.√3

AB = 20 - AB.√3

AB + AB.√3 = 20

AB.(1 + √3) = 20

AB = 20/(1 + √3)

AB = [20.(1 - √3)/(1 + √3).(1 - √3)]

AB = [20.(1 - √3)/(1 - 3)]

AB = 20.(1 - √3)/- 2

AB = 10.(√3 - 1)

→ AB ≈ 7.32 m

Recall: AB = BC.√2

BC = AB/√2 → where: AB = 10.(√3 - 1)

BC = 10.(√3 - 1)/√2

BC = 10.(√2).(√3 - 1)/2

BC = 5.(√6 - √2)

BC ≈ 5.17 m