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EJERCICIOS DE
APLICACIÓN LEY SENO
A
B
C
a
c
b
𝐴 = 180 − 60° − 75° = 45°
Aplicar ley de senos
𝑎
𝑠𝑒𝑛 𝐴
=
𝑏
𝑠𝑒𝑛 𝐵
=
𝑐
𝑠𝑒𝑛 𝐶
340
𝑠𝑒𝑛 45°
=
𝑏
𝑠𝑒𝑛 75°
=
𝑐
𝑠𝑒𝑛 60°
340
𝑠𝑒𝑛 45°
=
𝑐
𝑠𝑒𝑛 60°
480.83 =
𝑐
0.8660
480.83 0.8660 = 𝑐
𝑐 = 416.41 𝑘𝑚
APLICACIÓN DE LEY SENO
A= 87°
B= 84.5°
C=8.5°
b
a
𝑏
𝑠𝑒𝑛 84.5
=
500
𝑠𝑒𝑛 8.5°
𝑏
0.9953
= 3382.7
𝑏 = 3382.7 0.9953
= 3366.8
El teorema de la tangente relaciona las longitudes de dos lados de un triángulo con
las tangentes de los dos ángulos opuestos a éstos. Éste enuncia que:
La razón entre la suma de dos lados (a, b o c) de un triángulo y
su resta es igual a la razón entre la tangente de la media de los
dos ángulos opuestos a dichos lados y la tangente de la mitad de
la diferencia de éstos.
Aplicaciones del teorema de la tangente
•Sabiendo un lado y dos ángulos, calcular los otros dos lados y el otro ángulo.
•Sabiendo dos lados y un ángulo, calcular el otro lado y los otros dos ángulos.
LAA, LLA
Sea un triángulo con dos lados conocidos (a=4,2 cm y b=3,8 cm) y un ángulo conocido
(C=60º). Se desea hallar A, B y c.
Sabiendo que la suma de los ángulos de un triángulo son 180º:
𝐴 + 𝐵 + 𝐶 = 180°
𝐴 + 𝐵 = 180 ° − 𝐶 = 180 − 60 = 120°
4.2 + 3.8
4.2 − 3.8
=
tan
120
2
tan
𝐴 − 𝐵
2
20 =
1.7321
𝑡𝑎𝑛
𝐴 − 𝐵
2
tan
𝐴 − 𝐵
2
=
1.7321
20
tan
𝐴 − 𝐵
2
= 0.0866
𝐴 − 𝐵
2
= 𝑡𝑎𝑛−1
(0.0866)
𝐴 − 𝐵
2
= 4.9495
𝐴 − 𝐵 = (4.9495)(2)
𝐴 − 𝐵 = 9.899°
A=9.899+B
𝐴 + 𝐵 = 120
(9.899 + 𝐵) + 𝐵 = 120
2𝐵 = 120 − 9.899
2𝐵 = 110.101
𝐵 =
110.101
2
= 55.05°
A+B=120°
A=120-B = 120-55.05= 64.95°
POR LO TANTO LOS
ÁNGULOS SON:
A=64.95°
B=55.05°
C=60°
LAS MEDIDAS DE LOS
LADOS SON:
a=4.2
b=3.8
C=4.01 se calculó por ley
de senos
En el triángulo ABC, c = 10 cm, A = 68° , C = 36°, Resuelve el triángulo
𝑎 + 10
𝑎 − 10
=
tan
68 + 36
2
tan
68 − 36
2
𝑎 + 10
𝑎 − 10
=
tan(52)
tan(16)
𝑎 + 10
𝑎 − 10
=
1.2799
0.2867
𝑎 + 10
𝑎 − 10
= 4.4642
𝑎 + 10 = (𝑎 − 10)(4.4642)
𝑎 + 10 = 4.4642𝑎 − 44.642
-3.4642a=-54.642
𝑎 =
−54.642
−3.4642
𝑎 = 15.77
RESULTADOS:
A=68°
B= 76°
C= 36°
a=15.77
b= 16. 52 por ley seno
c= 10

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Ley tangente (02.12.2020)

  • 2. A B C a c b 𝐴 = 180 − 60° − 75° = 45° Aplicar ley de senos 𝑎 𝑠𝑒𝑛 𝐴 = 𝑏 𝑠𝑒𝑛 𝐵 = 𝑐 𝑠𝑒𝑛 𝐶 340 𝑠𝑒𝑛 45° = 𝑏 𝑠𝑒𝑛 75° = 𝑐 𝑠𝑒𝑛 60° 340 𝑠𝑒𝑛 45° = 𝑐 𝑠𝑒𝑛 60° 480.83 = 𝑐 0.8660 480.83 0.8660 = 𝑐 𝑐 = 416.41 𝑘𝑚
  • 3. APLICACIÓN DE LEY SENO A= 87° B= 84.5° C=8.5° b a 𝑏 𝑠𝑒𝑛 84.5 = 500 𝑠𝑒𝑛 8.5° 𝑏 0.9953 = 3382.7 𝑏 = 3382.7 0.9953 = 3366.8
  • 4.
  • 5. El teorema de la tangente relaciona las longitudes de dos lados de un triángulo con las tangentes de los dos ángulos opuestos a éstos. Éste enuncia que: La razón entre la suma de dos lados (a, b o c) de un triángulo y su resta es igual a la razón entre la tangente de la media de los dos ángulos opuestos a dichos lados y la tangente de la mitad de la diferencia de éstos.
  • 6. Aplicaciones del teorema de la tangente •Sabiendo un lado y dos ángulos, calcular los otros dos lados y el otro ángulo. •Sabiendo dos lados y un ángulo, calcular el otro lado y los otros dos ángulos. LAA, LLA
  • 7. Sea un triángulo con dos lados conocidos (a=4,2 cm y b=3,8 cm) y un ángulo conocido (C=60º). Se desea hallar A, B y c. Sabiendo que la suma de los ángulos de un triángulo son 180º: 𝐴 + 𝐵 + 𝐶 = 180° 𝐴 + 𝐵 = 180 ° − 𝐶 = 180 − 60 = 120° 4.2 + 3.8 4.2 − 3.8 = tan 120 2 tan 𝐴 − 𝐵 2 20 = 1.7321 𝑡𝑎𝑛 𝐴 − 𝐵 2 tan 𝐴 − 𝐵 2 = 1.7321 20
  • 8. tan 𝐴 − 𝐵 2 = 0.0866 𝐴 − 𝐵 2 = 𝑡𝑎𝑛−1 (0.0866) 𝐴 − 𝐵 2 = 4.9495 𝐴 − 𝐵 = (4.9495)(2) 𝐴 − 𝐵 = 9.899° A=9.899+B 𝐴 + 𝐵 = 120 (9.899 + 𝐵) + 𝐵 = 120 2𝐵 = 120 − 9.899 2𝐵 = 110.101 𝐵 = 110.101 2 = 55.05° A+B=120° A=120-B = 120-55.05= 64.95° POR LO TANTO LOS ÁNGULOS SON: A=64.95° B=55.05° C=60° LAS MEDIDAS DE LOS LADOS SON: a=4.2 b=3.8 C=4.01 se calculó por ley de senos
  • 9. En el triángulo ABC, c = 10 cm, A = 68° , C = 36°, Resuelve el triángulo 𝑎 + 10 𝑎 − 10 = tan 68 + 36 2 tan 68 − 36 2 𝑎 + 10 𝑎 − 10 = tan(52) tan(16) 𝑎 + 10 𝑎 − 10 = 1.2799 0.2867 𝑎 + 10 𝑎 − 10 = 4.4642 𝑎 + 10 = (𝑎 − 10)(4.4642)
  • 10. 𝑎 + 10 = 4.4642𝑎 − 44.642 -3.4642a=-54.642 𝑎 = −54.642 −3.4642 𝑎 = 15.77 RESULTADOS: A=68° B= 76° C= 36° a=15.77 b= 16. 52 por ley seno c= 10