# 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:

$A = 15+\dfrac{ 28 }{ 60 } = \dfrac{ 232 }{ 15 } \doteq 15.4667 \ ^\circ \ \\ B = 45 \ ^\circ \ \\ C = 180-A-B = 180-15.4667-45 = \dfrac{ 1793 }{ 15 } \doteq 119.5333 = 119.533 ^\circ = 119^\circ 32'$

Try calculation via our triangle calculator.

$S = 501.9 \ \\ a = k \cdot \ u \ \\ b = k \cdot \ v \ \\ c = k \cdot \ w \ \\ \ \\ w = 1 \ \\ u:w = \sin A: \sin C \ \\ u = w \cdot \ \sin(A) / \sin(C) = 1 \cdot \ \sin(15.4667^\circ ) / \sin(119.5333^\circ ) \doteq 0.3065 \ \\ \ \\ v:w = \sin B: \sin C \ \\ v = w \cdot \ \sin(B) / \sin(C) = 1 \cdot \ \sin(45^\circ ) / \sin(119.5333^\circ ) \doteq 0.8127 \ \\ \ \\ s = (u+v+w)/2 = (0.3065+0.8127+1)/2 = 1.0596 \ \\ T = \sqrt{ s \cdot \ (s-u) \cdot \ (s-v) \cdot \ (s-w) } = \sqrt{ 1.0596 \cdot \ (1.0596-0.3065) \cdot \ (1.0596-0.8127) \cdot \ (1.0596-1) } \doteq 0.1084 \ \\ k = \sqrt{ S/T } = \sqrt{ 501.9/0.1084 } \doteq 68.0562 \ \\ a = k \cdot \ u = 68.0562 \cdot \ 0.3065 \doteq 20.8593 = 20.859$
$b = k \cdot \ v = 68.0562 \cdot \ 0.8127 \doteq 55.3092 = 55.309$
$c = k \cdot \ w = 68.0562 \cdot \ 1 \doteq 68.0562 = 68.056$

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