I'm gonna label the corners of the outer square with A,B,C and D to make it easier, with A being the top left corner. And the corners of the inner rectangle with E, F, G, H with E being the upper most corner. [My apologies, all these letters may seem confusing to memorise so I suggest writing them down on a diagram somewhere so the solution is a little easier to follow]
Let angle AED = theta. From the larger unshaded triangle, we can tell that tantheta = 2.
Since AED, HEF, and BEF are supplementary, they add to 180, and HEF equals 90 whilst AED is theta. So doing the algebra (I trust that you can do this part since this question is one where if you can see it, the maths is pretty simple), and using angle sum of triangle, EFB equals theta. Using the same proof (supplementary angles, angle sum of triangle) angle FGC is also theta. Using the tantheta identity in each of the unshaded white triangles, you can get tantheta = something. (For the upper triangle, 1/2 is already given to you as a side length.) Let the other adjacent side be x, and let
1/2 over x = 2. (From the tantheta=2 identity you proved earlier on) And solve to get 1/4. Now since its a square the side length must be 1, so 1-1/4 gives 3/4 as the opposite angle in tantheta=2 for the bottom one.
(Oh, just a quick reminder that tantheta = opposite/adjacent, so the fraction itself can equal 2. Don't put tan(fraction)=2!)
Repeat the process for the bottom one, and you should get 3/8. Then, using pythagoreas theorem in each of the triangles, you should get respectively, rt(5)/4 and 3rt(5)/8. (Remember to square root the squares! It's a common silly error
)
Then multiply them. I got 15/32 which is between 7/16(7/16=14/32) and 1/2. (1/2=16/32)
