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I have a right triangle:

  • Height: y (value over 0)
  • Width: y (value over 0)
  • Angle: α (degrees, value between 0-90)

Right triangle

I need to find out the formula to count the length of x.

W0lfw00ds
  • 137

2 Answers2

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Yes, it's possible. The law of sines used on the lower triangle (the one spanned by $\alpha$) say that $$ \frac{\sin(135^\circ - \alpha)}{1} = \frac{\sin \alpha}{x} $$ where $135^\circ - \alpha$ is the measure of the top angle of the triangle (the lower left angle has measure $45^\circ$). So we have $$ x = \frac{\sin{\alpha}}{\sin(135^\circ - \alpha)} $$

Arthur
  • 199,419
  • Thanks for the answer. I need to check it. My friend offered his formula:

    x = ((x*Tan(a)) / (1+Tan(a))) / Sin(45) Is this also correct?

     – W0lfw00ds Apr 23 '14 at 08:14
  • Not as you've written it (note the $x$ on both sides of your equals sign: they will cancel and you're left with $1 = \dfrac{\frac{\tan\alpha}{1 + \tan\alpha}}{\sin 45^\circ}$ which is not true). – Arthur Apr 23 '14 at 08:17
  • Sorry. I had to change the width and height. I updated the question. I'll try to use your upper formula, by changing the value of 1 to match the height/width. – W0lfw00ds Apr 23 '14 at 08:53
  • The upper formula did the trick, when changing the "1" to y (to match the updated question). I used this as the final formula:

    x = Sin(a) / (Sin(135-a) / y)

     – W0lfw00ds Apr 23 '14 at 12:20
1

enter image description here

Using picture above, if $AB=BC=y$ then using Pythagoras' formula $AC=y\sqrt2$ and $CD=y\sqrt2-x$. Hence, using sine rule on triangle $ABD$ we get \begin{align} \frac{x}{\sin\alpha}&=\frac{y}{\sin(135^\circ-\alpha)}\\ x&=\frac{y\sin\alpha}{\sin(135^\circ-\alpha)} \end{align} or using sine rule on triangle $BCD$ we get \begin{align} \frac{y\sqrt2-x}{\sin(90^\circ-\alpha)}&=\frac{y}{\sin(45^\circ+\alpha)}\\ y\sqrt2-x&=\frac{y\sin(90^\circ-\alpha)}{\sin(45^\circ+\alpha)}\\ x&=y\sqrt2-\frac{y\sin(90^\circ-\alpha)}{\sin(45^\circ+\alpha)}\\ &=y\sqrt2\left(1-\frac{\cos\alpha}{\sin\alpha+\cos\alpha}\right)\\ &=\frac{y\sqrt2\sin\alpha}{\sin\alpha+\cos\alpha} \end{align}

Anastasiya-Romanova 秀
  • 19,345