Can you find area of triangle QST? | (Fun Geometry Problem) |

[@murdock5537]

Nice, but weird construction: PSQ = 0,4°→ QPS = 89,6°

[@DiffgaFysgie]

According to cosine law: PV² = 8² + 12² - 2*8*12*cos(R) => PV = 4*sqrt(13 - sqrt(7)).
According to sine law: 12/sin(P) = PV/sin(R) => sin(P) = 9/(4*sqrt(13 - sqrt(7)) => P ~ 44,366°, not 89,6°.
Where is my mistake?

[@murdock5537]

@@DiffgaFysgie Maybe this could be helpful:
φ = 30°; sin⁡(3φ) = 1; ∆ RPV → RP = RQ + PQ = (8 - a) + a
RV = RT + VT = (12 - b) + b; PV = PS + VS; RS = m; QS = 3; TS = 4
sin⁡(SQP) = sin⁡(VTS) = 1; QT = t
area ∆ PSR = (3/2)8 = 12; area ∆ CRS = (4/2)12 = 24 → PS = k = (1/2)VS → PV = 3k
area ∆ RPV = 36; SRQ = δ; TRS = γ → VRP = θ = δ + γ →
QST = 6φ - θ → sin⁡(6φ - θ) = sin⁡(θ)
∆ RPV → (1/2)sin⁡(θ)8(12) = 36 → sin⁡(θ) = 3/4 →
area ∆ QST= (1/2)sin⁡(θ)3(4) = 9/2; cos⁡(θ) = √7/4
∆ RPV → 9k^2 = 64 + 144 - 2(8)(12)cos⁡(θ) → k = (4/3)√(13 - 3√7)
PSQ = τ → cos⁡(τ) = 3/k = (9√106/106)√(13 + 3√7) →
arccos⁡(τ) ≈ 0,399447325° ≈ 0,4° = PSQ → QPS = 89,6°

[@marioalb9726]

A₁= ½.a.c + ½.b.d = ½(8*3+12*4)
A₁= 36 cm² = ½.a.b.sinα
sin α = 2*36/(8*12)= 3/4 = sin β
A = ½.c.d.sinβ = ½*3*4*3/4
A = 4,5 cm² ( Solved √ )
Same as vídeo !!!