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.
Calcula las razones trigonomΓ©tricas seno, coseno y tangente de los Γ‘ngulos agudos (A y B) del triΓ‘ngulo rectΓ‘ngulo que aparecen abajo.
Donde el valor de a o el
lado del Γ‘ngulo que falta la
medida se obtiene por
medio del teorema de
PitΓ‘goras.
h= πŸ’. πŸ“π’„π’Ž
a= 2.06 cm
b=4 cm
π’‰πŸ
= π’‚πŸ
+ π’ƒπŸ
π’‰πŸ
βˆ’ π’ƒπŸ
= π’‚πŸ
π’‚πŸ
= π’‰πŸ
βˆ’ π’ƒπŸ
π’‚πŸ
= (πŸ’. πŸ“π’„π’Ž)𝟐
βˆ’(πŸ’π’„π’Ž)𝟐
π’‚πŸ
= 𝟐𝟎. πŸπŸ“π’„π’ŽπŸ
βˆ’ πŸπŸ”π’„π’ŽπŸ
π’‚πŸ
= πŸ’. πŸπŸ“π’„π’ŽπŸ
π’‚πŸ
= πŸ’. πŸπŸ“π’„π’Ž
𝒂 = πŸ’. πŸπŸ“π’„π’Ž = π’„π’Ž
𝒂 =2.06m
1
2
.
β„Ž = 4.5π‘π‘š
π‘π‘œ = 2.06 π‘π‘š
π‘π‘Ž = 4π‘π‘š
π‘ π‘’π‘›π‘œ =
π‘π‘œ
β„Ž
πΆπ‘œπ‘ π‘’π‘›π‘œ =
π‘π‘Ž
β„Ž
π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘’ =
π‘π‘œ
π‘π‘Ž
se cancela seno a la menos 1 por seno y asΓ­ mismo con arco
coseno y seno, igualmente para tangente y arco tangente.
π‘ π‘’π‘›π‘œβˆ’1
π‘ π‘’π‘›π‘œ 𝐴 = π‘ π‘’π‘›π‘œβˆ’1
(
2.06π‘π‘š
4.5π‘π‘š
)
𝐴 = π‘ π‘’π‘›π‘œβˆ’1
2.06π‘π‘š
4.5π‘π‘š
𝑨 = πŸπŸ•. πŸπŸ’Β°
πΆπ‘œπ‘ βˆ’1
πΆπ‘œπ‘  𝐴 = πΆπ‘œπ‘ βˆ’1
4π‘π‘š
4.5π‘π‘š
𝐴 = πΆπ‘œπ‘ βˆ’1
4𝐢𝑀
4.5𝐢𝑀
𝑨 = πŸπŸ•. πŸπŸ’Β°
π‘‘π‘Žπ‘›π‘”βˆ’1
π‘‘π‘Žπ‘›π‘” 𝐴 = π‘‘π‘Žπ‘›π‘”βˆ’1
2.06π‘π‘š
4π‘π‘š
𝐴 = π‘‘π‘Žπ‘›π‘”βˆ’1
2.06π‘π‘š
4π‘π‘š
𝑨 = πŸπŸ•. πŸπŸ’Β°
h= πŸ’. πŸ“π’„π’Ž
a= 2.06 cm
b=4 cm
Una vez encontrado el valor del los tres lados
de los triΓ‘ngulos rectΓ‘ngulos para el Angulo
𝜢 π’š 𝜷 se procede a buscar el seno coseno
tangente.
En primer lugar se empezara con el Angulo A
y posteriormente 𝑩.
β„Ž = 4.5 π‘π‘š
π‘π‘œ = 4π‘π‘š
π‘π‘Ž = 2.06 π‘π‘š
π‘ π‘’π‘›π‘œ =
π‘π‘œ
β„Ž
πΆπ‘œπ‘ π‘’π‘›π‘œ =
π‘π‘Ž
β„Ž
π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘’ =
π‘π‘œ
π‘π‘Ž
π‘ π‘’π‘›π‘œ 𝐡 =
π‘π‘œ
β„Ž
π‘ π‘’π‘›π‘œβˆ’1
π‘ π‘’π‘›π‘œ 𝐡 = π‘ π‘’π‘›π‘œβˆ’1
4π‘π‘š
4.5π‘π‘š
𝐡 = π‘ π‘’π‘›π‘œβˆ’1
4π‘π‘š
4.5π‘π‘š
𝑩 = πŸ”πŸ. πŸ•πŸ‘Β°
πΆπ‘œπ‘ βˆ’1
πΆπ‘œπ‘  𝐡 = πΆπ‘œπ‘ βˆ’1
2.06π‘π‘š
4.5π‘π‘š
𝐡 = πΆπ‘œπ‘ βˆ’1
2.06π‘π‘š
4.5π‘π‘š
𝑩 = πŸ”πŸ, πŸ•πŸ‘Β°
π‘‘π‘Žπ‘›π‘”βˆ’1
π‘‘π‘Žπ‘›π‘” 𝐡 = π‘‘π‘Žπ‘›π‘”βˆ’1
4π‘π‘š
2,06π‘π‘š
𝐡 = π‘‘π‘Žπ‘›π‘”βˆ’1
4π‘π‘š
2,06π‘π‘š
𝑩 = πŸ”πŸ. πŸ•πŸ‘Β°
1
2
h= πŸ’. πŸ“π’„π’Ž
a= 2.06 cm
b=4 cm
β„Ž = 4.5 π‘π‘š
π‘π‘œ = 4π‘π‘š
π‘π‘Ž = 2.06 π‘π‘š
Una vez encontrado el Angulo A se
realiza la parte b donde encuentra el
valor del Angulo B.para realizar y
obtener el resultado de seno, coseno y
tangente para ello se debe tener
encuentra los valores de la
hipotenusa, cateto opuesto y cateto
adyacente
Β‘GRACIAS POR SU ATENCIΓ“N!

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Paso 3 tarea 2 a ejercicio grupal_unidad 2

  • 1. .
  • 2. Calcula las razones trigonomΓ©tricas seno, coseno y tangente de los Γ‘ngulos agudos (A y B) del triΓ‘ngulo rectΓ‘ngulo que aparecen abajo. Donde el valor de a o el lado del Γ‘ngulo que falta la medida se obtiene por medio del teorema de PitΓ‘goras. h= πŸ’. πŸ“π’„π’Ž a= 2.06 cm b=4 cm π’‰πŸ = π’‚πŸ + π’ƒπŸ π’‰πŸ βˆ’ π’ƒπŸ = π’‚πŸ π’‚πŸ = π’‰πŸ βˆ’ π’ƒπŸ π’‚πŸ = (πŸ’. πŸ“π’„π’Ž)𝟐 βˆ’(πŸ’π’„π’Ž)𝟐 π’‚πŸ = 𝟐𝟎. πŸπŸ“π’„π’ŽπŸ βˆ’ πŸπŸ”π’„π’ŽπŸ π’‚πŸ = πŸ’. πŸπŸ“π’„π’ŽπŸ π’‚πŸ = πŸ’. πŸπŸ“π’„π’Ž 𝒂 = πŸ’. πŸπŸ“π’„π’Ž = π’„π’Ž 𝒂 =2.06m
  • 3. 1 2 . β„Ž = 4.5π‘π‘š π‘π‘œ = 2.06 π‘π‘š π‘π‘Ž = 4π‘π‘š π‘ π‘’π‘›π‘œ = π‘π‘œ β„Ž πΆπ‘œπ‘ π‘’π‘›π‘œ = π‘π‘Ž β„Ž π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘’ = π‘π‘œ π‘π‘Ž se cancela seno a la menos 1 por seno y asΓ­ mismo con arco coseno y seno, igualmente para tangente y arco tangente. π‘ π‘’π‘›π‘œβˆ’1 π‘ π‘’π‘›π‘œ 𝐴 = π‘ π‘’π‘›π‘œβˆ’1 ( 2.06π‘π‘š 4.5π‘π‘š ) 𝐴 = π‘ π‘’π‘›π‘œβˆ’1 2.06π‘π‘š 4.5π‘π‘š 𝑨 = πŸπŸ•. πŸπŸ’Β° πΆπ‘œπ‘ βˆ’1 πΆπ‘œπ‘  𝐴 = πΆπ‘œπ‘ βˆ’1 4π‘π‘š 4.5π‘π‘š 𝐴 = πΆπ‘œπ‘ βˆ’1 4𝐢𝑀 4.5𝐢𝑀 𝑨 = πŸπŸ•. πŸπŸ’Β° π‘‘π‘Žπ‘›π‘”βˆ’1 π‘‘π‘Žπ‘›π‘” 𝐴 = π‘‘π‘Žπ‘›π‘”βˆ’1 2.06π‘π‘š 4π‘π‘š 𝐴 = π‘‘π‘Žπ‘›π‘”βˆ’1 2.06π‘π‘š 4π‘π‘š 𝑨 = πŸπŸ•. πŸπŸ’Β° h= πŸ’. πŸ“π’„π’Ž a= 2.06 cm b=4 cm Una vez encontrado el valor del los tres lados de los triΓ‘ngulos rectΓ‘ngulos para el Angulo 𝜢 π’š 𝜷 se procede a buscar el seno coseno tangente. En primer lugar se empezara con el Angulo A y posteriormente 𝑩.
  • 4. β„Ž = 4.5 π‘π‘š π‘π‘œ = 4π‘π‘š π‘π‘Ž = 2.06 π‘π‘š π‘ π‘’π‘›π‘œ = π‘π‘œ β„Ž πΆπ‘œπ‘ π‘’π‘›π‘œ = π‘π‘Ž β„Ž π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘π‘’ = π‘π‘œ π‘π‘Ž π‘ π‘’π‘›π‘œ 𝐡 = π‘π‘œ β„Ž π‘ π‘’π‘›π‘œβˆ’1 π‘ π‘’π‘›π‘œ 𝐡 = π‘ π‘’π‘›π‘œβˆ’1 4π‘π‘š 4.5π‘π‘š 𝐡 = π‘ π‘’π‘›π‘œβˆ’1 4π‘π‘š 4.5π‘π‘š 𝑩 = πŸ”πŸ. πŸ•πŸ‘Β° πΆπ‘œπ‘ βˆ’1 πΆπ‘œπ‘  𝐡 = πΆπ‘œπ‘ βˆ’1 2.06π‘π‘š 4.5π‘π‘š 𝐡 = πΆπ‘œπ‘ βˆ’1 2.06π‘π‘š 4.5π‘π‘š 𝑩 = πŸ”πŸ, πŸ•πŸ‘Β° π‘‘π‘Žπ‘›π‘”βˆ’1 π‘‘π‘Žπ‘›π‘” 𝐡 = π‘‘π‘Žπ‘›π‘”βˆ’1 4π‘π‘š 2,06π‘π‘š 𝐡 = π‘‘π‘Žπ‘›π‘”βˆ’1 4π‘π‘š 2,06π‘π‘š 𝑩 = πŸ”πŸ. πŸ•πŸ‘Β° 1 2 h= πŸ’. πŸ“π’„π’Ž a= 2.06 cm b=4 cm β„Ž = 4.5 π‘π‘š π‘π‘œ = 4π‘π‘š π‘π‘Ž = 2.06 π‘π‘š Una vez encontrado el Angulo A se realiza la parte b donde encuentra el valor del Angulo B.para realizar y obtener el resultado de seno, coseno y tangente para ello se debe tener encuentra los valores de la hipotenusa, cateto opuesto y cateto adyacente
  • 5. Β‘GRACIAS POR SU ATENCIΓ“N!