Question

ayuda por favorrrrrrrrrr............

Answer 1

Respuesta:

1) k = 10

2) J = 49/100

Explicación paso a paso:

1) 0,333... equivale a 1/3

senA = co/h = 1/3

co= 1

h = 3

calculamos el cateto adyacente con el teorema de pitagoras

$c^{2}=h^{2}-c^{2}$

$h^{2}=3^{2}-1^{2}$

$h^{2}=9-1$

$h^{2}=8$

h = $\sqrt{8}$

h = 2$\sqrt{2}$

cosA = (2$\sqrt{2}$)3

tanC = (2$\sqrt{2})$/1 = $2\sqrt{2}$

cotA = (2$\sqrt{2})/1=2\sqrt{2}$

k = ($cos^{2}A+tan^{2}C)/(sen^{2}A+cot^{2}A)$

k = ((2$\sqrt{2}/3)^{2}+(2\sqrt{2})^{2})/((1/3)^{2}.(2\sqrt{2})^{2})$

k = ((4.2/9) + (4.2))/((1/9)(4.2))

k = (8/9 + 8)/((1/9)(8))

k = (80/9)/(8/9)

k = 720/72

k = 10

2)

cscB = h/co = 10/7

h = 10

co = 7

calculamos el otro cateto por el teorema de pitagoras

$c^{2}=10^{2}-7^{2}$

$c^{2}$ = 100 - 49

$c^{2}$ = 51

C = $\sqrt{51}$

senA = $\sqrt{51}/10$

cosA = 7/10

tanB = 7/$\sqrt{51}$.$\sqrt{51}/\sqrt{51}=7\sqrt{51}/51$

J = senAcosAtanB

J = $\sqrt{51}/10$.7/10.7$\sqrt{51}/51$

J = 49$\sqrt{2601}$/5100

J = (49.51)/5100

J = 2499/5100 = 833/1700 = 49/100