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An isosceles triangle with two sides $\tau = 2\pi$ (one is the base), altitude 1, and area $\pi$. What is the length of third side and the three angles (in both radians and degrees)?

Note: not a right triangle, so values are not the same as Archimedes's triangle for the area of a disk.

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    $\begingroup$ What have you tried? You have the area. Imagine bisecting the angle between the two $2\pi$ sides to get two right triangles. $\endgroup$ – Ross Millikan May 19 at 23:08
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Say your triangle is $ABC$ with $AB=AC=\tau$ and the altitude $BH=1$, where $H$ in $AC$. You can use pythagorean theorem to get $AH=\sqrt{AB^2-BH^2}=\sqrt{\tau^2-1^2}$. Then, $HC=AC-AH=\tau-\sqrt{\tau^2-1^2}$. Finally, use pythagorean theorem again to compute that $$BC=\sqrt{BH^2+HC^2}=\sqrt{1+(\tau-\sqrt{\tau^2-1^2})^2}$$.

To obtain the angles, you express the angles in terms of inverse trig functions, for instance $\angle A=\arcsin(BH/AB)=\arcsin(1/\tau)$, and so on.

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  • $\begingroup$ The last two angles can be done using: $\angle B = \angle C$ and $ (\pi - \arcsin(1/\tau))/2$. Thanks. Only one real use of $\pi$ nice and clean with $\tau$. $\endgroup$ – John Nicholson May 20 at 0:35
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Hints:

  • Find $AD$ by Pythagoras
  • Find $DC$ by subtraction
  • Find $BC$ by Pythagoras
  • Find angles by trigonometry

enter image description here

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  • $\begingroup$ how did you draw that $\endgroup$ – endgame yourgame May 20 at 3:10
  • $\begingroup$ is this geogebra $\endgroup$ – endgame yourgame May 20 at 3:10
  • $\begingroup$ @endgameendgame: MSPaint - it is not to scale $\endgroup$ – Henry May 20 at 7:55

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