Given A B = 12.65 m , B C = 20.0 m , α = 18 ° 3 2 ′ 1 5 ′ ′ . AB=12.65m, BC=20.0m, \alpha=18\degree32'15''. A B = 12.65 m , BC = 20.0 m , α = 18°3 2 ′ 1 5 ′′ .
β = 90 ° − α \beta=90\degree-\alpha β = 90° − α Then
∠ A B C = β = 90 ° − α \angle ABC=\beta=90\degree-\alpha ∠ A BC = β = 90° − α Triangle ACB
A B = 12.65 , B C = 20 , ′ ′ AB=12.65, BC=20, '' A B = 12.65 , BC = 20 , ′′
∠ A B C = 90 ° − α = 90 ° − 18 ° 3 2 ′ 1 5 ′ ′ \angle ABC=90\degree-\alpha=90\degree-18\degree32'15'' ∠ A BC = 90° − α = 90° − 18°3 2 ′ 1 5 ′′ a) Law of Cosines
A C 2 = A B 2 + B C 2 − 2 ( A B ) ( B C ) cos ∠ A B C {AC}^2={AB}^2+{BC}^2-2(AB)(BC)\cos\angle ABC A C 2 = A B 2 + BC 2 − 2 ( A B ) ( BC ) cos ∠ A BC
A C = 12.65 2 + 20 2 − 2 ( 12.65 ) ( 20 ) sin ( 18 ° 3 2 ′ 1 5 ′ ′ ) AC=\sqrt{{12.65}^2+{20}^2-2(12.65)(20)\sin(18\degree32'15'')} A C = 12.65 2 + 20 2 − 2 ( 12.65 ) ( 20 ) sin ( 18°3 2 ′ 1 5 ′′ )
A C = 19.98 m AC=19.98m A C = 19.98 m 2) Law of Sines
sin ∠ A C B A B = sin ∠ A B C A C \dfrac{\sin\angle ACB}{AB}=\dfrac{\sin \angle ABC}{AC} A B sin ∠ A CB = A C sin ∠ A BC
∠ A C B = sin − 1 ( A B sin ∠ A B C A C ) \angle ACB=\sin^{-1}(\dfrac{AB\sin \angle ABC}{AC}) ∠ A CB = sin − 1 ( A C A B sin ∠ A BC )
∠ A C B = sin − 1 ( 12.65 cos ( 18 ° 3 2 ′ 1 5 ′ ′ ) 19.98 ) \angle ACB=\sin^{-1}(\dfrac{12.65\cos (18\degree32'15'')}{19.98}) ∠ A CB = sin − 1 ( 19.98 12.65 cos ( 18°3 2 ′ 1 5 ′′ ) )
∠ A C B = 36.89 ° \angle ACB=36.89\degree ∠ A CB = 36.89°
∠ A C B = 36 ° 5 3 ′ 2 5 ′ ′ \angle ACB=36\degree53'25'' ∠ A CB = 36°5 3 ′ 2 5 ′′
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