Question

determinar las 6 razones trigonometricas del angulo"A" de un triangulo rectangulo ABC,recto en "B",si se sabe que 5a=b

Answer 1

Hola! \ (•◡•) /

Graficaremos el trángulo rectángulo(Ver imagen), luego haremos lo siguiente

$5a=b\\ \\ \frac{a}{b}=\frac{1}{5}\\ \\ Entonces \\ a = k\\ b= 5k$

Hallaremos el cateto faltante con el Teorema de Pitágoras

$b^{2} = a^{2}+c^{2} \\ \\ (5k)^{2} =k^{2} +c^{2} \\ \\ 25k^{2} =k^{2} +c^{2}\\ \\ 24k^{2} =c^{2}\\ \\ c =\sqrt{24k^{2}} \\ \\ c = 2\sqrt{6} k$

Las razones trigonométricas serán:

1. Para el ángulo A

• $senA = \frac{a}{b} = \frac{k}{5k} = \frac{1}{5}$
• $cos A = \frac{c}{b} = \frac{2\sqrt{6}k}{5k}= \frac{2\sqrt{6}}{5}$
• $tan A = \frac{a}{c} = \frac{k}{2\sqrt{6}k}= \frac{1}{2\sqrt{6}}$
• $cot A = \frac{c}{a} = \frac{2\sqrt{6}k}{k}= 2\sqrt{6}$
• $sec A = \frac{b}{c} = \frac{5k}{2\sqrt{6}k}= \frac{5}{2\sqrt{6}}$
• $cscA = \frac{b}{a} = \frac{5k}{k} = 5$

2. Para el ángulo C

• $sen C = \frac{c}{b} = \frac{2\sqrt{6}k}{5k}= \frac{2\sqrt{6}}{5}$
• $cosC = \frac{a}{b} = \frac{k}{5k} = \frac{1}{5}$
• $tan C = \frac{c}{a} = \frac{2\sqrt{6}k}{k}= 2\sqrt{6}$
• $cot C = \frac{a}{c} = \frac{k}{2\sqrt{6}k}= \frac{1}{2\sqrt{6}}$
• $sec C = \frac{b}{a} = \frac{5k}{k} = 5$
• $csc C = \frac{b}{c} = \frac{5k}{2\sqrt{6}k}= \frac{5}{2\sqrt{6}}$