another geometry
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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
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
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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
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
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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.
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.