Tunnel - quadrilateral

How long will tunnel AB be, given distances AD = 35 m, DC = 120 m, CB = 85 m, angle ADC = 105°, and angle BCD = 71°, where ABCD is a quadrilateral?

Final Answer:

a = 111.5661 m

d = 35 m c = 120 m b = 85 m δ = 105^{\circ} γ = 71^{\circ} A = (0, 0) D = (x_{0}, y_{0}) φ_{1} = 0^{\circ} x_{0} = A_{x} + d \cdot cos (φ_{1}) = 0 + 35 \cdot cos 0 = 35 y_{0} = A_{y} + d \cdot sin (φ_{1}) = 0 + 35 \cdot sin 0 = 0 C = (x_{1}, y_{1}) φ_{2} = (180 - δ) + φ_{1} = (180 - 105) + 0 = 75^{\circ} x_{1} = x_{0} + c \cdot cos φ_{2} = x_{0} + c \cdot cos 75 ° = 35 + 120 \cdot cos 75 ° = 35 + 120 \cdot 0.258819 = 66.05829 y_{1} = y_{0} + c \cdot sin φ_{2} = y_{0} + c \cdot sin 75 ° = 0 + 120 \cdot sin 75 ° = 0 + 120 \cdot 0.965926 = 115.9111 B = (x_{2}, y_{2}) φ_{3} = (180 - γ) + φ_{2} = (180 - 71) + 75 = 184^{\circ} x_{2} = x_{1} + b \cdot cos φ_{3} = x_{1} + b \cdot cos 184 ° = 66.0583 + 85 \cdot cos 184 ° = 66.0583 + 85 \cdot (- 0.997564) = - 18.73466 y_{2} = y_{1} + b \cdot sin φ_{3} = y_{1} + b \cdot sin 184 ° = 115.9111 + 85 \cdot sin 184 ° = 115.9111 + 85 \cdot (- 0.069756) = 109.9818 a = ∣ A B ∣ a = (x_{2} - A_{x})^{2} + (y_{2} - A_{y})^{2} = ((- 18.7347) - 0)^{2} + (109.9818 - 0)^{2} = 111.5661 m

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