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 interior angles α = 15°28' and β = 45°.

Final Answer:

C =  119.5333 °
a =  20.8592
b =  55.309
c =  68.0557

Step-by-step explanation:

$$\begin{gathered} A=15°28^{\prime }=15°+\frac{28}{60}°=15.4667°\doteq 15.4667 \\ B=45{}^{\circ } \\ C=180-A-B=180-15.4667-45=119.5333^{\circ }=119°32^{\prime } \end{gathered}$$

Try calculation via our triangle calculator.

$$\begin{gathered} S=501.9 \\ a=k\cdot u \\ b=k\cdot v \\ c=k\cdot w \\ w=1 \\ u:w=\sin A:\sin C \\ u=w\cdot \sin (A)/\sin (C)=1\cdot \sin (15.4667°)/\sin (119.5333°)\doteq 0.3065 \\ v:w=\sin B:\sin C \\ v=w\cdot \sin (B)/\sin (C)=1\cdot \sin (45°)/\sin (119.5333°)\doteq 0.8127 \\ s=(u+v+w)/2=(0.3065+0.8127+1)/2\doteq 1.0596 \\ T=\sqrt{s\cdot (s-u)\cdot (s-v)\cdot (s-w)}=\sqrt{1.0596\cdot (1.0596-0.3065)\cdot (1.0596-0.8127)\cdot (1.0596-1)}\doteq 0.1084 \\ k=\sqrt{S/T}=\sqrt{501.9/0.1084}\doteq 68.0557 \\ a=k\cdot u=68.0557\cdot 0.3065=20.8592 \end{gathered}$$

$b=k\cdot v=68.0557\cdot 0.8127=55.309$

$c=k\cdot w=68.0557\cdot 1=68.0557$

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