Question

A triangle has two corners with angles of `( pi ) / 3` and `( pi )/ 6`. If one side of the triangle has a length of `1`, what is the largest possible area of the triangle?

Answer 1

If the hypotenuse is 1 then `Area_(max)=root2(3)/8`
If one of the two catheti is 1 then `Area_(max)=roots(3)/2`

Explanation:

In any case we are dealing with a rectangle triangle.

If the hypotenuse is 1, the cathetus `x` is the base whereas the cathetus `y` is the height and its area is `Area=x*y/2` under the constraint that `x^2+y^2=1`. If `y` is the cathetus opposite to the `pi/3` angle, from trigonometry we know that `y=x*tan(pi/3)=xroot2(3)`. As a result the area is `Area=x^2root2(3)/2`. But we can deduce `x^2` from the constraint and it results `3x^2+x^2=1` from which we have `x^2=1/4`. Replaced this in the area formula, finally we have `Area=root2(3)/8`

If the base is 1 then `y=root2(3)` and `Area=1*y/2=root2(3)/2`