sábado, 21 de mayo de 2011

LEY DE LOS SENOS

Ley de los senos


Encontrar todos los elementos de un triangulo oblicuángulo y su superficie.

A=70°
a=36
B=48°
b=28.47
C= 62°
c=33.82
S=452.34us²

A+B+C=180°
70°+48°+C=180°
118°+C=180°
C=180°-118°
C=62°










































A= bh/2
Sen = c.o/h
Sen 62°= h/28.47
Sen 62°(28.47)=h
H= 25.13
S= 36(25.13)/2
S=452.34us²
















2-

A= 63°
a=16.5cm
B=47°
b=13.5434cm
C=70°
c=17.4015cm
S=104.9675us²

A+B+C=180°
A+47°+70°=180°
A+117°=180°
A=180°-117°
C=63°










































A= bh/2
Sen = c.o/h

Sen 63°= h/13.54 cm
Sen 63°(13.54 cm)=h
H= 12.0642 cm
S= 17.4015 cm(12.0642 cm )/2
S=104.9675 us²














3

A= 17°
a=46.0115cm
B=77°
b=153.34cm
C=86°
c=157cm
S=3519.155us²


A+B+C=180°
17°+77°+C=180°
94°+C=180°
C=180°-94°
C=86°


































A= bh/2
Sen = c.o/h
Sen 17°= h/153.34 cm
Sen 17°(153.34 cm)=h
H= 44.83 cm
S= 157 cm(44.83 cm )/2
S=3519.155 us²














4


A= 70°
a=85.87mts
B=80°
b=90mts
C=30°
c=45.69 mts
S=1,932.075 us²

A+B+C=180°
70°+B+ 30°=180°
B+100°=180°
B=180°-100°
B=80°


































A= bh/2
Sen = c.o/h
Sen 30°= h/85.87 mts
Sen 30°(85.87 mts)=h
H= 42.935 mts
S= 90 mts (42.935)/2
S=1,932.075 us²
















5


A= 83°
a=21cm
B=47°
b=15.47 cm
C=50°
c=16.20 cm
S=124.425 us²

A+B+C=180°
A+47°+50°=180°
A+97°=180°
A=180°-97°
C=83°






































A= bh/2
Sen = c.o/h
Sen 50°= h/15.47 cm
Sen 50°(15.47 cm)=h
H= 11.85 cm
S= 21 cm(11.85 cm )/2
S=124.425 us²




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