calculated sinx right triangles in whom the difference cathetus c sqrt2/2  c is the hypotenuse

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember the Pythagorean theorem in right angle triangles such that:

`c^2 = a^2 + b^2`

c represents the hypotenuse

a,b represent the legs of right angle triangle

The problem provides the information that the difference of legs gives `c*sqrt2/2`  such that:

`a - b = c*sqrt2/2`

Raising to square both sides yields:

`a^2 - 2ab + b^2 = c*sqrt2/2`

Substituting `c^2`  for  `a^2 + b^2`  yields:

`c^2 - 2ab = c*sqrt2/2 => 2ab = c^2 - c*sqrt2/2`

`a^2 + b^2 = (a+b)^2 - 2ab => (a+b)^2 - 2ab = c^2`

`(a+b)^2 = c^2 + c^2 - c*sqrt2/2 => a+b = +-sqrt(2c^2 - c*sqrt2/2)`

You may use Lagrange's resolvents such that:

`x^2 - (a+b)x + ab = 0`

`Delta = (a+b)^2 - 4ab = 2c^2 - c*sqrt2/2 - 2c^2 + c*sqrt2`

`Delta = c*sqrt2/2 => sqrt Delta = +-sqrt(c*sqrt2/2)`

`a = (sqrt(2c^2 - c*sqrt2/2)+sqrt(c*sqrt2/2))/2`

`b= (sqrt(2c^2 - c*sqrt2/2)-sqrt(c*sqrt2/2))/2`

Hence, since the legs a and b are expressed in terms of c, you may evaluate sine function in the given right angle triangle, such that:

`sin alpha = a/c => sin alpha = (sqrt(2c^2 - c*sqrt2/2)+sqrt(c*sqrt2/2))/(2c)`

`sin beta = b/c => sin beta = (sqrt(2c^2 - c*sqrt2/2)-sqrt(c*sqrt2/2))/(2c)`

Hence, evaluating the values of sine function in the given right angle triangle yields`sin alpha = (sqrt(2c^2 - c*sqrt2/2)+sqrt(c*sqrt2/2))/(2c)`  and `sin beta = (sqrt(2c^2 - c*sqrt2/2)-sqrt(c*sqrt2/2))/(2c).`

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