I have a right angled triangle in which I know the length and the slope of the hypotenuse, how do I find one of the angles? Thanks

  • $\begingroup$ Could you explain fully please? $\endgroup$ – Jerry Murphy Dec 9 '14 at 15:08
  • $\begingroup$ Knowing the slope of the hypothenuse allows you to express one of the side in terms of the other one. Then, you can relate those two sides to the length of the hypothenuse (pythagoras theorem) and you will have an equation with only one unknown. $\endgroup$ – Olivier Dec 9 '14 at 15:15
  • $\begingroup$ One of the angles is $90^\circ$ :) $\endgroup$ – peterwhy Dec 9 '14 at 15:35
  • $\begingroup$ While I'm aware that a right angled triangle has an angle of 90 degrees that's about the limit off my knowledge...it's been 15 years since I've done any maths. I don't know how to express one of the sides in terms of the other one. However if someone could answer the question, in future I should be able to answer similar ones. $\endgroup$ – Jerry Murphy Dec 9 '14 at 15:56
  • $\begingroup$ Using the end points of the “hypotenuse” of known length as diameter, create a circle through them. Any one point on the circumference, together with those two end points, can be your right angled triangle. This means there are infinitely many answers and no unique solution to your question. Knowing the slope of the hypotenuse does not help. $\endgroup$ – Mick Dec 9 '14 at 16:13

$$\text{Slope} = \dfrac{\text{height}}{\text{base}}=\tan(\ \text{Base}\angle\ )$$

You know the base length and hypotenuse. Then:

$$\text{height} = \sqrt{\text{hypotenuse}^2-\text{base}^2}$$

Putting the value of $height$ in the first equation, we get $$\dfrac{\sqrt{\text{hypotenuse}^2-\text{base}^2}}{\text{base}}=\tan(\ \text{Base}\angle\ )$$

$$\implies\ \text{Base}\ \angle=\arctan\left(\dfrac{\sqrt{\text{hypotenuse}^2-\text{base}^2}}{\text{base}}\right)$$


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