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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. πππ a= 2.06 cm b=4 cm ππ = ππ + ππ ππ β ππ = ππ ππ = ππ β ππ ππ = (π. πππ)π β(πππ)π ππ = ππ. πππππ β πππππ ππ = π. πππππ ππ = π. ππππ π = π. ππππ = ππ π =2.06m
3.
1 2 . β = 4.5ππ ππ
= 2.06 ππ ππ = 4ππ π πππ = ππ β πΆππ πππ = ππ β π‘ππππππ‘π = ππ ππ se cancela seno a la menos 1 por seno y asΓ mismo con arco coseno y seno, igualmente para tangente y arco tangente. π πππβ1 π πππ π΄ = π πππβ1 ( 2.06ππ 4.5ππ ) π΄ = π πππβ1 2.06ππ 4.5ππ π¨ = ππ. ππΒ° πΆππ β1 πΆππ π΄ = πΆππ β1 4ππ 4.5ππ π΄ = πΆππ β1 4πΆπ 4.5πΆπ π¨ = ππ. ππΒ° π‘πππβ1 π‘πππ π΄ = π‘πππβ1 2.06ππ 4ππ π΄ = π‘πππβ1 2.06ππ 4ππ π¨ = ππ. ππΒ° h= π. πππ 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 πΆ π π· se procede a buscar el seno coseno tangente. En primer lugar se empezara con el Angulo A y posteriormente π©.
4.
β = 4.5
ππ ππ = 4ππ ππ = 2.06 ππ π πππ = ππ β πΆππ πππ = ππ β π‘ππππππ‘π = ππ ππ π πππ π΅ = ππ β π πππβ1 π πππ π΅ = π πππβ1 4ππ 4.5ππ π΅ = π πππβ1 4ππ 4.5ππ π© = ππ. ππΒ° πΆππ β1 πΆππ π΅ = πΆππ β1 2.06ππ 4.5ππ π΅ = πΆππ β1 2.06ππ 4.5ππ π© = ππ, ππΒ° π‘πππβ1 π‘πππ π΅ = π‘πππβ1 4ππ 2,06ππ π΅ = π‘πππβ1 4ππ 2,06ππ π© = ππ. ππΒ° 1 2 h= π. πππ a= 2.06 cm b=4 cm β = 4.5 ππ ππ = 4ππ ππ = 2.06 ππ 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
ATENCIΓN!
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