# Isosceles triangle has two sides of length $2\pi$ and area $\pi$. Find the third side and the three angles.

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.

• What have you tried? You have the area. Imagine bisecting the angle between the two $2\pi$ sides to get two right triangles. May 19 '20 at 23:08

Hints:

• Find $$AD$$ by Pythagoras
• Find $$DC$$ by subtraction
• Find $$BC$$ by Pythagoras
• Find angles by trigonometry • how did you draw that May 20 '20 at 3:10
• is this geogebra May 20 '20 at 3:10
• @endgameendgame: MSPaint - it is not to scale May 20 '20 at 7:55

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.

• 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 $\tau$. May 20 '20 at 0:35