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In a right angle triangle, base = `sqrt(x^(2)+8x)`, height = 4. What is the hypotenuse...

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user354582 | (Level 1) Honors

Posted June 22, 2013 at 4:00 AM via web

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In a right angle triangle, base = `sqrt(x^(2)+8x)`, height = 4. What is the hypotenuse and the angle `theta` . Also, what are the trignometric functions of `theta` in terms of x.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 22, 2013 at 4:09 AM (Answer #1)

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The base of the right triangle is `sqrt(x^2 + 8x)` . The height is 4. The length of the hypotenuse in terms of x is `sqrt((sqrt(x^2 + 8x))^2+ 16)` = `sqrt(x^2 + 8x + 16)` = `sqrt((x+4)^2)` = `x + 4`

The angle `theta = sin^-1(4/(x+4))`

`cos theta = sqrt(x^2 + 8x)/(x+4)`

`tan theta = 4/sqrt(x^2 + 8x)`

For the other values use : `cot theta = 1/tan theta` , `sec theta = 1/cos theta` and `cosec theta = 1/sin theta` .

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