if there is a right triangle with the hypotenuse being 5 units and the long leg being 4 uits, what would the formula be to find angle x between the ypotenuse and the shorter leg?

Guest Feb 17, 2012

#1**0 **

5^2=4^2+x^2

5^2-4^2=x^2

x=sqrt(5^2-4^2)

x=sqrt(25-16)

x=sqrt(9)

x=3 (shorter leg length)

cos( alpha ) = x/hypotenuse

alpha = acos( x/hypotenuse )

alpha = acos( 3/5 )

[input]acos( 3/5 )[/input]

apersonwithaquestion:if there is a right triangle with the hypotenuse being 5 units and the long leg being 4 uits, what would the formula be to find angle x between the ypotenuse and the shorter leg?

5^2=4^2+x^2

5^2-4^2=x^2

x=sqrt(5^2-4^2)

x=sqrt(25-16)

x=sqrt(9)

x=3 (shorter leg length)

cos( alpha ) = x/hypotenuse

alpha = acos( x/hypotenuse )

alpha = acos( 3/5 )

[input]acos( 3/5 )[/input]

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Feb 17, 2012