Question

doy 80 puntosssssssssssssssssssss​

Answer 1

Respuesta:

no

Explicación paso a paso:

Answer 2

Respuesta:

Explicación paso a paso:

a

$x=\sqrt{9^2-5^2}=2\sqrt{14}\\\\senA=cosB=\frac{2\sqrt{14}}{9}\\\\senB=cosA=\frac{5}{9}\\\\tanA=\frac{2\sqrt{14}}{5}\\\\tanB=\frac{5\sqrt{14}}{28}$

b

$x=\sqrt{3^2-(\sqrt5)^2}=2\\\\senA=cosB=\frac{2}{3}\\\\senB=cosA=\frac{\sqrt5}{3}\\\\tanA=\frac{2\sqrt{5}}{5}\\\\tanB=\frac{\sqrt5}{2}$

c

$x=\sqrt{10^2+7^2}=\sqrt{149}\\\\senN=cosM=\frac{7\sqrt{149}}{149}\\\\senM=cosN=\frac{10\sqrt{149}}{149}\\\\tanN=\frac{7}{10}\\\\tanM=\frac{10}{7}$

d

$x=\sqrt{(\sqrt2)^2-1^2}=1\\\\senA=cosB=senB=cosA=\frac{\sqrt2}{2}\\\\tanA=tanB=1$

a

$cos\theta=sen\alpha=\frac{1}{5}\\\\sen\theta=cos\alpha\approx0,980\\\\tan\theta\approx4,899\\\\tan\alpha\approx0,204$

b

$senA=cosB\approx0,832\\\\senB=cosA\approx0,555\\\\tanA=\frac{3}{2}\\\\tanB=\frac{2}{3}$

c

$senN=cosM=\frac{1}{2}\\\\senM=cosN=\frac{\sqrt3}{2}\\\\tanA=\frac{\sqrt{3}}{3}\\\\tanB=\sqrt{3}$

d

$sen\alpha=cos\theta=\frac{\sqrt{3}}{6}\\\\sen\theta=cos\alpha\approx0,957\\\\tan\alpha\approx0,302\\\\tan\theta\approx3,317$

e

$sen\alpha=cos\beta\approx0,791\\\\sen\beta=cos\alpha\approx0,612\\\\tan\alpha=\frac{\sqrt{15}}{3}\\\\tan\beta=\frac{\sqrt{15}}{5}$

f

$senA=cosB=\frac{4\sqrt{29}}{29}\\\\senB=cosA\approx0,670\\\\tanA\approx1,109\\\\tanB\approx0,901$