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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)**