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by vikrantr93 » Mon May 02, 2011 3:59 am
2 persons start walking on a road that diverge at an angle of 120 degree. if they walk at rate 3 km/h and 2 km/h respectively. Find the distance between them after 4 hours.

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a. 5 km
2. 4√19 km
c. 7 km
d. 8√19 km
e. None of these
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by manpsingh87 » Mon May 02, 2011 4:21 am
vikrantr93 wrote:2 persons start walking on a road that diverge at an angle of 120 degree. if they walk at rate 3 km/h and 2 km/h respectively. Find the distance between them after 4 hours.

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a. 5 km
2. 4√19 km
c. 7 km
d. 8√19 km
e. None of these
CosC= a^2+b^2-c^2/2ab;
here angle C=120; a=12; b=8;c=?? (as two persons are traveling at the speed of 3 and 2km/h, therefore distance covered in 4 hours will be 3*4=12 and 2*4=8)
Cos 120=-1/2;
-1/2=144+64-c^2/2 12*8;
c^2=304;
c=4sqrt(19) hence [spoiler]B;[/spoiler]
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by Brent@GMATPrepNow » Mon May 02, 2011 9:45 am
vikrantr93 wrote:2 persons start walking on a road that diverge at an angle of 120 degree. if they walk at rate 3 km/h and 2 km/h respectively. Find the distance between them after 4 hours.

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a. 5 km
2. 4√19 km
c. 7 km
d. 8√19 km
e. None of these
Please note that the above solution uses the Cosine Law.
You don't need to know this law for the GMAT. In fact, you don't need to know anything about sines, cosines or tangents for the GMAT.

My solution requires the addition of some lines and knowledge of 30-60-90 "special" triangles (which is required for the GMAT).

Cheers,
Brent

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by Ashishkapoor7 » Sun Jun 03, 2012 8:13 am
Awesome Brent! your course Rocks!!
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by dhonu121 » Thu Jun 07, 2012 8:10 pm
After 4 hours, one has travelled 12 km and the other has 8 kms.
Now if we do a little reconstrunction and drop a perpendicular from A on BC extended, we have a right angled triangle, with perpedicular, BC extended and AB as hipotenuse. AB is needed now.
length of perpedicular = AC sin 60 = 12sin60 = 6sqrt3
Length of BC extended = 8+ACcos60 = 8 + 4 = 12.
Now length of AB = Length of hipotensuse.
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