ANS C

Stmt 1 - Length of AC is given and from the questions 2 angles are given so we can calculate

tan 45 = AC/BC - Calculate BC From here

tan 60 = AC/DC - Calculate DC from here

BC-DC = BD... SUFF

Stmt 2 - BA/DA is given

sin 45 = AC/AB ----- (1)

sin 60 = AC/AD ----- (2)

Divide (1) by (2) and AC can be calculated

once AC is known you can calculate BD as stated for stmt 1 SUFF

But answer is A

the simulator does not gives any explanation so cant find its logic.

Hmmmm... After reevaluating I realized that I was wrong when I said

Stmt 2 - BA/DA is given

sin 45 = AC/AB ----- (1)

sin 60 = AC/AD ----- (2)

Divide (1) by (2) and AC can be calculated


Actually AC cannot be calculated as this is commin in NUM and DEN... so will cancel each other...

STUPID MISTAKE... Ans is A

Ans is A

A) sufficient you can find BC and DC and hence BD
B) not sufficient because only ratios are provided

BA/DA = sqrt(6)/2
BA = DA x sqrt(6)/2
BC = BA/sqrt(2) = DA x sqrt(3)/2
DC = DA/2

so BC-DC = DA (sqrt(3)/2-1/2)

hence difference will be in terms of DA or BA hence insufficient.