Area and two angles

Calculate the size of all sides and internal angles of a triangle ABC, if it is given by area S = 501.9; and two internal angles α = 15°28' and β = 45°.

Result

C =  119.533 °
a =  20.859
b =  55.309
c =  68.056

Solution:

A=15+2860=2321515.4667  B=45  C=180AB=18015.466745=179315119.5333=119.533=11932A = 15+\dfrac{ 28 }{ 60 } = \dfrac{ 232 }{ 15 } \doteq 15.4667 \ ^\circ \ \\ B = 45 \ ^\circ \ \\ C = 180-A-B = 180-15.4667-45 = \dfrac{ 1793 }{ 15 } \doteq 119.5333 = 119.533 ^\circ = 119^\circ 32'

Try calculation via our triangle calculator.

S=501.9 a=k u b=k v c=k w  w=1 u:w=sinA:sinC u=w sin(A)/sin(C)=1 sin(15.4667)/sin(119.5333)0.3065  v:w=sinB:sinC v=w sin(B)/sin(C)=1 sin(45)/sin(119.5333)0.8127  s=(u+v+w)/2=(0.3065+0.8127+1)/2=1.0596 T=s (su) (sv) (sw)=1.0596 (1.05960.3065) (1.05960.8127) (1.05961)0.1084 k=S/T=501.9/0.108468.0562 a=k u=68.0562 0.306520.8593=20.859S = 501.9 \ \\ a = k \cdot \ u \ \\ b = k \cdot \ v \ \\ c = k \cdot \ w \ \\ \ \\ w = 1 \ \\ u:w = \sin A: \sin C \ \\ u = w \cdot \ \sin(A) / \sin(C) = 1 \cdot \ \sin(15.4667^\circ ) / \sin(119.5333^\circ ) \doteq 0.3065 \ \\ \ \\ v:w = \sin B: \sin C \ \\ v = w \cdot \ \sin(B) / \sin(C) = 1 \cdot \ \sin(45^\circ ) / \sin(119.5333^\circ ) \doteq 0.8127 \ \\ \ \\ s = (u+v+w)/2 = (0.3065+0.8127+1)/2 = 1.0596 \ \\ T = \sqrt{ s \cdot \ (s-u) \cdot \ (s-v) \cdot \ (s-w) } = \sqrt{ 1.0596 \cdot \ (1.0596-0.3065) \cdot \ (1.0596-0.8127) \cdot \ (1.0596-1) } \doteq 0.1084 \ \\ k = \sqrt{ S/T } = \sqrt{ 501.9/0.1084 } \doteq 68.0562 \ \\ a = k \cdot \ u = 68.0562 \cdot \ 0.3065 \doteq 20.8593 = 20.859
b=k v=68.0562 0.812755.3092=55.309b = k \cdot \ v = 68.0562 \cdot \ 0.8127 \doteq 55.3092 = 55.309
c=k w=68.0562 168.0562=68.056c = k \cdot \ w = 68.0562 \cdot \ 1 \doteq 68.0562 = 68.056







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