adi_800 wrote:If CD = 6, what is the length of BC?
(1) BD = 6 rt3
(2) x = 60
great question, IMO E
Length of BC
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great question, IMO E
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sanju09
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adi_800 wrote:If CD = 6, what is the length of BC?
(1) BD = 6 rt3
(2) x = 60
[1] In triangle BCD, CD = 6, DB = 6 √3, and angle BDC = 30º. Now, come what may, BC could also be found out uniquely. Sufficient
[2] In triangle BCD, CD = 6, angle BDC = 30º, and angle BCD = 120º. Now, come what may, BC could also be found out uniquely. Sufficient
[spoiler]D[/spoiler]
Last edited by sanju09 on Mon Sep 20, 2010 1:09 am, edited 1 time in total.
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sumit.sinha
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adi_800 wrote:If CD = 6, what is the length of BC?
(1) BD = 6 rt3
(2) x = 60
My Explanation:
Draw a perpendicular line from B to AC, which will intersect line AC at E.
Let BC = a, EC = b and EB = c
(1)
in the triangle BED, it is a 30:60:90 triangle. Hence the sides are in the ratio 1x : xrt3 : 2x
as 2x = 6rt3.
therefore, x = 3 rt3
Hence, EB = c = x = 3rt3
EC = b = ED - CD = xrt3 - 6 = (3 rt 3 * rt3) - 6 = 9-6 = 3
Now in the triangle BEC, using Pythagoras theorem,
a^2 = b^2 + c^2
i.e a = 6
SUFFICIENT
(2)
In the triangle BEC, It is a 30:60:90 triangle. Hence the sides are in the ratio 1x : xrt3 : 2x
as 2x = a
therefore x = a/2
b will become, b =a/2
and c =(a/2) rt3
Now in triangle BED, again it is a 30:60:90 triangle. Hence the sides are in the ratio 1y : yrt3 : 2y [ taking variable y so as not to get confused between the earlier x and this y]
as side ED is opposite to angle 60. therefore, yrt3 = (a/2) + 6
i.e. y = (((a/2)+6)/ rt3)
Also as side EB is opposite to angle 30, so y = (a/2) rt 3
equating both ys,
(((a/2)+6)/ rt3) = (a/2) rt 3
solving, we get a =6
SUFFICIENT
Therefore CORRECT ANSWER (D)
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akhpad
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Aditya Bhai
This is not the difficult problem. I can select D without solving.
Statement 1:
you know two sides and included angle; so you can find other sides and angles by applying cosine or sine formula.
Statement 2:
angle BDC = angle CBD = 30 degree
DC = BC = 6
This is not the difficult problem. I can select D without solving.
Statement 1:
you know two sides and included angle; so you can find other sides and angles by applying cosine or sine formula.
Statement 2:
angle BDC = angle CBD = 30 degree
DC = BC = 6
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anantbhatia
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Need a clarification over here.
(1) definitely only one way to draw a triangle with the given two sides and included angle. So surely, one can determine the length by drawing the triangle and measuring its side. Is that sufficient? Or does sufficiency means that the required has to be derived mathematically?
(2) same as above. Sufficient by drawing, but mathematically not sure.
Can some expert answer this query? It is more of a basic doubt about DS rather than about this question.
(1) definitely only one way to draw a triangle with the given two sides and included angle. So surely, one can determine the length by drawing the triangle and measuring its side. Is that sufficient? Or does sufficiency means that the required has to be derived mathematically?
(2) same as above. Sufficient by drawing, but mathematically not sure.
Can some expert answer this query? It is more of a basic doubt about DS rather than about this question.
In the question is X = 60 supposed to be the measure of the angle BCA or length of the side AC?
Its not clear... if its the angle measure - which is what I thought it was - then (2) is easily sufficient -- cause then you can prove that the triangle is isoceles..
Statement 1 also seems sufficient as 2 sides and included angle are fixed...
Its not clear... if its the angle measure - which is what I thought it was - then (2) is easily sufficient -- cause then you can prove that the triangle is isoceles..
Statement 1 also seems sufficient as 2 sides and included angle are fixed...
