Question

ayudenmeeee por favorrrr

Answer 1

Para angulo q
Senq=2/3 = 0.6666
q = arcsen(0.6666)= 0.7297 radianes
$K= \sqrt{5}* tan(0.7297)+ \frac{1}{ \sqrt{5} } cos(0.7297)= \\ K=2.236*0.894427+0.4472*0.74535= \\ K=2+0.33= \\ K=2.3333=2 \frac{1}{3}= \frac{7}{3}$
Respuesta inciso d) 7/3

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$Siendo\hspace{2mm} Ctg\theta=Cos37grados$
$Cos 37grados=0.798636=Ctg\theta \\ Ctg\theta= \frac{1}{tan\theta} \\ \\ tan\theta= \frac{1}{ctg\theta} = \frac{1}{0.798636} =1.2521 \\ \\ \Theta=Atan(1.2521)= 0.8969 \hspace{2mm}radianes$

$E=(Csc ^{2} \theta-1)Tg\theta \\ Csc ^{2}=( \frac{1}{sen\theta} ) ^{2} =1.6378 \\ \\ Tg\theta=1.252\\ \\ E=(1.6378-1)*1.252=0.8 \\ \\ Solucion:\hspace{2mm} Inciso \hspace{2mm} d) 0.8$


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Calcular x , si :   $sen\theta= \frac{1}{2}$

$sen\theta= \frac{1}{2} = \frac{2x+1}{x+11} \\ 4x+2=x+11 \\ \\ 4x-x=11-2 \\ \\ 3x=9 \\ \\ \\ x= \frac{9}{3} =3 \\ \\ Inciso\hspace{2mm}a) 3$