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Let BD and AD be b and h respectively.
Area of triangle ABD = 2 sq m ( by symmetry)
So, 1/2 * b * h = 2
or b*h = 4
or b = 4/h ----------(1)
Also, b^2 + h^2 = 16 (pythagoras theorem) ------------------------(2)
On putting the value of (1) in (2) we get:
(4/h)^2 + h^2 =16
or (16 + h^4)/h2 = 16
or h^4 -16h^2 +16 = 0
Put x = h^2 and solve for x we get:
x = 8 - 4√3 or x = 8 + 4√3
So, h = 2√(2 - √3) or h = 2√(2 + √3) and neglecting other two negative values of h, since h cannot be negative.
Sin x = 2√(2 - √3)/4 = or sinx = 2√(2 + √3)/4, since sinx = perpendicular/hyp.
of sinx = √(2 - √3)/2 = or sinx = √(2 + √3)/2
if sinx = √(2 - √3)/2
then x = arcsin(√(2 - √3)/2) = <B =<C = 15 degrees
and <A = 180 -2*15 = 150 degrees
if sinx = √(2 + √3)/2 then
x = arcsin(√(2 + √3)/2) =<B=<C = 75 degrees
and <A = 180 - 2*75 = 30 degrees
So the angles of the two isosceles triangles are:
(15,15,150) and (75,75,30)