Sinus-Rechner

Löse das Dreieck, wenn $$$b = 3$$$, $$$A = 60^0$$$, $$$B = 45^0$$$.

Nach dem Sinusgesetz: $$$\frac{a}{\sin{\left(A \right)}} = \frac{b}{\sin{\left(B \right)}}$$$.

In unserem Fall $$$\frac{a}{\sin{\left(60^0 \right)}} = \frac{3}{\sin{\left(45^0 \right)}}$$$.

Also $$$a = \frac{3 \sin{\left(60^0 \right)}}{\sin{\left(45^0 \right)}} = \frac{3 \sqrt{6}}{2}$$$.

Der dritte Winkel ist $$$C = 180^0 - \left(A + B\right)$$$.

In unserem Fall $$$C = 180^0 - \left(60^0 + 45^0\right) = 75^0$$$.

Nach dem Sinusgesetz: $$$\frac{c}{\sin{\left(C \right)}} = \frac{b}{\sin{\left(B \right)}}$$$.

In unserem Fall $$$\frac{c}{\sin{\left(75^0 \right)}} = \frac{3}{\sin{\left(45^0 \right)}}$$$.

Also $$$c = \frac{3 \sin{\left(75^0 \right)}}{\sin{\left(45^0 \right)}} = \frac{3 \left(1 + \sqrt{3}\right)}{2}$$$.

Das Gebiet ist $$$S = \frac{1}{2} a b \sin{\left(C \right)} = \left(\frac{1}{2}\right)\cdot \left(\frac{3 \sqrt{6}}{2}\right)\cdot \left(3\right)\cdot \left(\sin{\left(75^0 \right)}\right) = \frac{9 \left(\sqrt{3} + 3\right)}{8}$$$

Der Umfang ist $$$P = a + b + c = \frac{3 \sqrt{6}}{2} + 3 + \frac{3 \left(1 + \sqrt{3}\right)}{2} = \frac{3 \left(\sqrt{3} + \sqrt{6} + 3\right)}{2}$$$

$$$a = \frac{3 \sqrt{6}}{2}\approx 3.674234614174767$$$A

$$$b = 3$$$A

$$$c = \frac{3 \left(1 + \sqrt{3}\right)}{2}\approx 4.098076211353316$$$A

$$$A = 60^0$$$A

$$$B = 45^0$$$A

$$$C = 75^0$$$A

Bereich: $$$S = \frac{9 \left(\sqrt{3} + 3\right)}{8}\approx 5.323557158514987$$$A.

Umfang: $$$P = \frac{3 \left(\sqrt{3} + \sqrt{6} + 3\right)}{2}\approx 10.772310825528083$$$A.