Tough Geometry Problem

This topic has expert replies
User avatar
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Thu Feb 03, 2011 11:51 pm
Thanked: 1 times

Tough Geometry Problem

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.

Image


a. 5 km
2. 4√19 km
c. 7 km
d. 8√19 km
e. None of these

User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 am
Thanked: 72 times
Followed by:6 members

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.

Image


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]
O Excellence... my search for you is on... you can be far.. but not beyond my reach!

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

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.

Image


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

Image
Brent Hanneson - Creator of GMATPrepNow.com
Image

User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Fri Jan 14, 2011 4:04 am
Thanked: 1 times

by Ashishkapoor7 » Sun Jun 03, 2012 8:13 am
Awesome Brent! your course Rocks!!

Master | Next Rank: 500 Posts
Posts: 316
Joined: Sun Aug 21, 2011 6:18 am
Thanked: 16 times
Followed by:6 members

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.
If you've liked my post, let me know by pressing the thanks button.