calcule sen,cos e tg dos angulos dados
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Soluções para a tarefa
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a)
ângulo α
seno = 3/5
cosseno = 4/5
tangente = 3/4
ângulo β
seno = 4/5
cosseno = 3/5
tangente = 4/3
b)
ângulo α
seno = 5/√50 = (5*√50) / (√50 *√50) = 5√50/50 = √50/10
cosseno = 5/√50 = (5*√50) / (√50 *√50) = 5√50/50 = √50/10
tangente = 5/5 = 1
c)
ângulo α
seno = 9/√192 = (9*√192) / (√192 * √192) = 9*√192/192 = 3√192/64
cosseno = 11/√192 = (11*√192) / (√192*√192) = 11*√192/192
tangente = 9/11
angulo β
seno = 11/√192 = (11*√192) / (√192*√192) = 11*√192/192
cosseno = 9/√192 = (9*√192) / (√192 * √192) = 9*√192/192 = 3√192/64
tangente = 11/9
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26/11/2015
SSRC - Sepauto
*-*-*-*-*-*-*-*-*-*-*-*
ângulo α
seno = 3/5
cosseno = 4/5
tangente = 3/4
ângulo β
seno = 4/5
cosseno = 3/5
tangente = 4/3
b)
ângulo α
seno = 5/√50 = (5*√50) / (√50 *√50) = 5√50/50 = √50/10
cosseno = 5/√50 = (5*√50) / (√50 *√50) = 5√50/50 = √50/10
tangente = 5/5 = 1
c)
ângulo α
seno = 9/√192 = (9*√192) / (√192 * √192) = 9*√192/192 = 3√192/64
cosseno = 11/√192 = (11*√192) / (√192*√192) = 11*√192/192
tangente = 9/11
angulo β
seno = 11/√192 = (11*√192) / (√192*√192) = 11*√192/192
cosseno = 9/√192 = (9*√192) / (√192 * √192) = 9*√192/192 = 3√192/64
tangente = 11/9
*-*-*-*-*-*-*-*-*-*-*-*
26/11/2015
SSRC - Sepauto
*-*-*-*-*-*-*-*-*-*-*-*
MoniBarbosa:
muito obrigada ;)
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