Question

Look at the image.

Answer 1

ans:39

Tap on the photo

first prove triangle BHC similar to FDC.Reason(AA)

then,provrSBH and GBH congruent(SAS)

CB=CG

CBG is an isolesis triangle

angle G and CBG is X

angle BCG=180-90-12=180-78=102

X=(108-102)/2=39

Answer 2

Answer:39°

Step-by-step explanation:

If you know some properties of circle you may solve it like this:

Since ∠CDF = 90°, FC is the diameter of the circle passing through C, D and F.

So B is the centre of the circle mentioned above as B is the mid-point of FC.

Then we have BF = BC = BD since they are all radii of the same circle.

∠BDF = ∠BFD = 12°

∠BDA = 90° + 12° = 102°

∠BDA + ∠BAD + ∠ABD = 180°

102° + x + x = 180°

x = (180° - 102°)/2 = 39°

To issacchoy1234: Nvm if you don't understand, this will be taught in 2nd term of S4 ;)