Kalkylator för sinussatsen

Lös triangeln, om $$$b = 3$$$, $$$A = 60^{\circ}$$$, $$$B = 45^{\circ}$$$.

Enligt sinussatsen: $$$\frac{a}{\sin{\left(A \right)}} = \frac{b}{\sin{\left(B \right)}}$$$.

I vårt fall gäller $$$\frac{a}{\sin{\left(60^{\circ} \right)}} = \frac{3}{\sin{\left(45^{\circ} \right)}}$$$.

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

Den tredje vinkeln är $$$C = 180^{\circ} - \left(A + B\right)$$$.

I vårt fall gäller $$$C = 180^{\circ} - \left(60^{\circ} + 45^{\circ}\right) = 75^{\circ}$$$.

Enligt sinussatsen: $$$\frac{c}{\sin{\left(C \right)}} = \frac{b}{\sin{\left(B \right)}}$$$.

I vårt fall gäller $$$\frac{c}{\sin{\left(75^{\circ} \right)}} = \frac{3}{\sin{\left(45^{\circ} \right)}}$$$.

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

Arean är $$$S = \frac{1}{2} a b \sin{\left(C \right)} = \left(\frac{1}{2}\right)\cdot \left(\frac{3 \sqrt{6}}{2}\right)\cdot \left(3\right)\cdot \left(\sin{\left(75^{\circ} \right)}\right) = \frac{9 \left(\sqrt{3} + 3\right)}{8}.$$$

Omkretsen är $$$P = a + b + c = \frac{3 \sqrt{6}}{2} + 3 + \frac{3 \left(1 + \sqrt{3}\right)}{2} = \frac{3 \left(\sqrt{3} + \sqrt{6} + 3\right)}{2}$$$.

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

$$$b = 3$$$A

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

$$$A = 60^{\circ}$$$A

$$$B = 45^{\circ}$$$A

$$$C = 75^{\circ}$$$A

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

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