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Resolver triángulo rectángulo si $$$b = 239000$$$, $$$B = 55^{\circ}$$$, $$$C = 90^{\circ}$$$
Tu aportación
Resuelve el triángulo, si $$$b = 239000$$$, $$$B = 55^{\circ}$$$, $$$C = 90^{\circ}$$$.
Solución
Según la definición del seno: $$$c = \frac{b}{\sin{\left(B \right)}}$$$.
Por lo tanto, $$$c = \frac{239000}{\sin{\left(\frac{11 \pi}{36} \right)}}$$$.
El tercer ángulo es $$$A = 180^{\circ} - \left(B + C\right)$$$.
En nuestro caso, $$$A = 180^{\circ} - \left(55^{\circ} + 90^{\circ}\right) = 35^{\circ}$$$.
Según la ley de los senos: $$$\frac{a}{\sin{\left(A \right)}} = \frac{b}{\sin{\left(B \right)}}$$$.
En nuestro caso, $$$\frac{a}{\sin{\left(35^{\circ} \right)}} = \frac{239000}{\sin{\left(55^{\circ} \right)}}$$$.
Por lo tanto, $$$a = \frac{239000 \sin{\left(35^{\circ} \right)}}{\sin{\left(55^{\circ} \right)}} = 239000 \tan{\left(\frac{7 \pi}{36} \right)}$$$.
El área es $$$S = \frac{1}{2} a b = \left(\frac{1}{2}\right)\cdot \left(239000 \tan{\left(\frac{7 \pi}{36} \right)}\right)\cdot \left(239000\right) = 28560500000 \tan{\left(\frac{7 \pi}{36} \right)}.$$$
El perímetro es $$$P = a + b + c = 239000 \tan{\left(\frac{7 \pi}{36} \right)} + 239000 + \frac{239000}{\sin{\left(\frac{11 \pi}{36} \right)}} = \frac{239000 \left(1 + \sqrt{2} \cos{\left(\frac{\pi}{18} \right)}\right)}{\sin{\left(\frac{11 \pi}{36} \right)}}.$$$
Respuesta
$$$a = 239000 \tan{\left(\frac{7 \pi}{36} \right)}\approx 167349.601632120637291$$$A
$$$b = 239000$$$A
$$$c = \frac{239000}{\sin{\left(\frac{11 \pi}{36} \right)}}\approx 291765.126713988000332$$$A
$$$A = 35^{\circ}$$$A
$$$B = 55^{\circ}$$$A
$$$C = 90^{\circ}$$$A
Área: $$$S = 28560500000 \tan{\left(\frac{7 \pi}{36} \right)}\approx 1.9998277395038416156225138 \cdot 10^{10}.$$$A
Perímetro: $$$P = \frac{239000 \left(1 + \sqrt{2} \cos{\left(\frac{\pi}{18} \right)}\right)}{\sin{\left(\frac{11 \pi}{36} \right)}}\approx 698114.728346108637623.$$$A