Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 May 2015, 09:52

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# solve without trig

Author Message
TAGS:
Manager
Joined: 14 Jun 2008
Posts: 161
Followers: 1

Kudos [?]: 26 [0], given: 0

solve without trig [#permalink]  19 Jul 2008, 15:26
hi,

any suggestions on how to use the following question without trig i.e. without using sines or cosines

cheers

S
Attachments

doubt Q1.JPG [ 48.91 KiB | Viewed 861 times ]

Director
Joined: 23 Sep 2007
Posts: 797
Followers: 5

Kudos [?]: 100 [0], given: 0

Re: solve without trig [#permalink]  19 Jul 2008, 15:29
draw the vertical lines from point P and point Q to the x axis. solve by using 30-60-90 degrees triangle values.
Current Student
Joined: 18 Jun 2008
Posts: 228
Followers: 2

Kudos [?]: 55 [0], given: 3

Re: solve without trig [#permalink]  19 Jul 2008, 15:32
using 30/60/90 triangle ratios can allow you to solve it,
its 1/2/sqrt3
30 degrees on the left hand side, 60 on the right hand side (because the middle is 90 degrees, and a semi circle has 180)

GMAT Instructor
Joined: 24 Jun 2008
Posts: 978
Location: Toronto
Followers: 281

Kudos [?]: 799 [0], given: 3

Re: solve without trig [#permalink]  19 Jul 2008, 16:22
Or notice the lines are perpendicular, so their slopes are negative reciprocals:

-from (-root(3), 1) to (0,0), you go right by root(3) and down by 1.

-from the origin, on a perpendicular line, to find a point the same distance away as (-root(3), 1), you can go right by 1 and up by root(3). The point is (1, root(3)).
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Director
Joined: 20 Sep 2006
Posts: 658
Followers: 2

Kudos [?]: 77 [0], given: 7

Re: solve without trig [#permalink]  19 Jul 2008, 16:35
Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.

20 sec solution:

2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).

Ans is D
SVP
Joined: 30 Apr 2008
Posts: 1892
Location: Oklahoma City
Schools: Hard Knocks
Followers: 34

Kudos [?]: 467 [0], given: 32

Re: solve without trig [#permalink]  19 Jul 2008, 16:43
rao_1857,

There are actually quite a few triangles made with this. I see what you're saying that the radius forms a triangle also, so it would be an iscoceles triangle, but I don't believe that is the easiest method for solving this problem.

We know point O is the origin, and then the point on the left is $$(\sqrt{3},1)$$ If you use the x-axis as one side of that triangle, and then go up to the point, you get another triangle. This one is actually a 30:60:90 triangle. The bottom (or x-axis) has a length of $$sqrt(3)$$ and the other side of the triangle (the hypotnuse is the same as the radius) is 1, so $$1^2 + sqrt{3}^2 = 4$$ so the radius is 2.

We don't even really need to know this to solve the problem. Because the middle angle formed by the two radii (?) is a 90 degree angle, we know that the values of the points for x and y will switch. x = $$sqrt{3}$$ so t on the other side = $$sqrt{3}$$ and y = 1, so s =1 on the other side. Actually a very simple problem and no trig needed.
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Current Student
Joined: 18 Jun 2008
Posts: 228
Followers: 2

Kudos [?]: 55 [0], given: 3

Re: solve without trig [#permalink]  19 Jul 2008, 17:10
rao_1857 wrote:
Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.

20 sec solution:

2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).

Ans is D

while you're right that the middle triangle is an isoceles triangle, your conclusion is flawed - the angles are not the same on either side of the middle isoceles triangle and so the distance does not remain the same.

Your reasoning would only be correct if the angle on either side of the middle triangle is equal - which is not the case here. So the answer is actually 1 - choice B, and not sqrt 3 - choice D
Director
Joined: 20 Sep 2006
Posts: 658
Followers: 2

Kudos [?]: 77 [0], given: 7

Re: solve without trig [#permalink]  19 Jul 2008, 18:45
jallenmorris wrote:
rao_1857,

There are actually quite a few triangles made with this. I see what you're saying that the radius forms a triangle also, so it would be an iscoceles triangle, but I don't believe that is the easiest method for solving this problem.

We know point O is the origin, and then the point on the left is $$(\sqrt{3},1)$$ If you use the x-axis as one side of that triangle, and then go up to the point, you get another triangle. This one is actually a 30:60:90 triangle. The bottom (or x-axis) has a length of $$sqrt(3)$$ and the other side of the triangle (the hypotnuse is the same as the radius) is 1, so $$1^2 + sqrt{3}^2 = 4$$ so the radius is 2.

We don't even really need to know this to solve the problem. Because the middle angle formed by the two radii (?) is a 90 degree angle, we know that the values of the points for x and y will switch. x = $$sqrt{3}$$ so t on the other side = $$sqrt{3}$$ and y = 1, so s =1 on the other side. Actually a very simple problem and no trig needed.

Thats right and I see your point but why do we need to consider that triangle if we can easily solve it using the isoceles traigle with 2 radii. I am little confused now....
Director
Joined: 20 Sep 2006
Posts: 658
Followers: 2

Kudos [?]: 77 [0], given: 7

Re: solve without trig [#permalink]  19 Jul 2008, 18:48
jasonc wrote:
rao_1857 wrote:
Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.

20 sec solution:

2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).

Ans is D

while you're right that the middle triangle is an isoceles triangle, your conclusion is flawed - the angles are not the same on either side of the middle isoceles triangle and so the distance does not remain the same.

Your reasoning would only be correct if the angle on either side of the middle triangle is equal - which is not the case here. So the answer is actually 1 - choice B, and not sqrt 3 - choice D

Pardon my ignorance. May be I am looking from diff view. But you are that angles are not same on the isocelec traingle..... by definition of isoceles triangle this is not true.
GMAT Instructor
Joined: 24 Jun 2008
Posts: 978
Location: Toronto
Followers: 281

Kudos [?]: 799 [0], given: 3

Re: solve without trig [#permalink]  19 Jul 2008, 21:57
rao_1857 wrote:
Guys dont confuse your self. This is a isoceles traingle. It is not 30,60,90 traingle.

20 sec solution:

2 sides are radius of same circle. So you have to make sure that the distance remains same (i.e. sqrt(3)). Since we know that the center is 0,0. S = sqrt(3).

Ans is D

If Q is any other point on the circle, you'll get an isosceles triangle. If Q is (0,2), or (2,0), or (root(2), root(2)), the distances OP and OQ will be equal. If you aren't somehow using the right angle in the diagram, you aren't doing the problem correctly.

The answer is 1, as explained above.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Director
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 209 [0], given: 0

Re: solve without trig [#permalink]  19 Jul 2008, 22:16
you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer $$sqrt(3)$$

the 30-60-90 traingle mentioned by JasonC are actually these two traingles.

Attachment:

circle.JPG [ 9.22 KiB | Viewed 770 times ]
GMAT Instructor
Joined: 24 Jun 2008
Posts: 978
Location: Toronto
Followers: 281

Kudos [?]: 799 [0], given: 3

Re: solve without trig [#permalink]  19 Jul 2008, 22:40
Nice diagram, durgesh, and a good illustration of what's really happening here (but your 30 degree angle is bigger than the 60 degree angle! ) The angle on the left should be 30, and on the right should be 60, with the given co-ordinates for P and Q.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

SVP
Joined: 29 Aug 2007
Posts: 2497
Followers: 57

Kudos [?]: 556 [0], given: 19

Re: solve without trig [#permalink]  19 Jul 2008, 22:42
durgesh79 wrote:
you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer $$sqrt(3)$$

the 30-60-90 traingle mentioned by JasonC are actually these two traingles.

Attachment:
circle.JPG

nice one, but 60 should be 30 and 30 should be 60.
_________________
Director
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 209 [0], given: 0

Re: solve without trig [#permalink]  19 Jul 2008, 22:48
GMAT TIGER wrote:
durgesh79 wrote:
you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer $$sqrt(3)$$

the 30-60-90 traingle mentioned by JasonC are actually these two traingles.

Attachment:
circle.JPG

nice one, but 60 should be 30 and 30 should be 60.

oops
Director
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 209 [0], given: 0

Re: solve without trig [#permalink]  19 Jul 2008, 22:58
IanStewart wrote:
Nice diagram, durgesh, and a good illustration of what's really happening here (but your 30 degree angle is bigger than the 60 degree angle! ) The angle on the left should be 30, and on the right should be 60, with the given co-ordinates for P and Q.

Thanks Ian... it was a typo ...
i think I now can distinguish between 60 and 30 angles.... so no need for me to go back to highschool geomatry books ....
Intern
Joined: 29 Jun 2008
Posts: 37
Followers: 0

Kudos [?]: 6 [0], given: 0

Re: solve without trig [#permalink]  20 Jul 2008, 05:23
Quote:
Thats right and I see your point but why do we need to consider that triangle if we can easily solve it using the isoceles traigle with 2 radii. I am little confused now....

Because the isoceles triangle with 2 radii is "tilted" - its hypotenuse is not parallel to the x-axis. If, instead, P was (-2/sqrt(2),2/sqrt(2)) then Q was (2/sqrt(2),2/sqrt(2))
Manager
Joined: 14 Jun 2008
Posts: 161
Followers: 1

Kudos [?]: 26 [0], given: 0

Re: solve without trig [#permalink]  20 Jul 2008, 06:06
jallenmorris wrote:
rao_1857,

There are actually quite a few triangles made with this. I see what you're saying that the radius forms a triangle also, so it would be an iscoceles triangle, but I don't believe that is the easiest method for solving this problem.

We know point O is the origin, and then the point on the left is $$(\sqrt{3},1)$$ If you use the x-axis as one side of that triangle, and then go up to the point, you get another triangle. This one is actually a 30:60:90 triangle. The bottom (or x-axis) has a length of $$sqrt(3)$$ and the other side of the triangle (the hypotnuse is the same as the radius) is 1, so $$1^2 + sqrt{3}^2 = 4$$ so the radius is 2.

We don't even really need to know this to solve the problem. Because the middle angle formed by the two radii (?) is a 90 degree angle, we know that the values of the points for x and y will switch. x = $$sqrt{3}$$ so t on the other side = $$sqrt{3}$$ and y = 1, so s =1 on the other side. Actually a very simple problem and no trig needed.

how do you know "the values of the points for x and y will switch"?
by trying values, i can see this, but is there a theorom for this?
Manager
Joined: 14 Jun 2008
Posts: 161
Followers: 1

Kudos [?]: 26 [0], given: 0

Re: solve without trig [#permalink]  20 Jul 2008, 06:08
so i guess i have to remember my cosine and sine ratios for 30, 45 and 60?
Director
Joined: 27 May 2008
Posts: 550
Followers: 7

Kudos [?]: 209 [0], given: 0

Re: solve without trig [#permalink]  20 Jul 2008, 06:16
sset009 wrote:
so i guess i have to remember my cosine and sine ratios for 30, 45 and 60?

if you are talking abt this question, i dont think so....

Slope of line OP = $$(1-0)/(-sqrt3 - 0) = - 1/sqrt3$$
slope of line OQ = t/s

both lines are perpendicular = $$(-1/sqrt3) * (t/s) = -1$$

$$t = sqrt3 * s$$ -------- (1)

both P and Q are points on circle
$$t^2 + s^2 = 1^2 + (sqrt3)^2 = 4$$ --------(2)

Solve for s and you'll get s = 1
Director
Joined: 20 Sep 2006
Posts: 658
Followers: 2

Kudos [?]: 77 [0], given: 7

Re: solve without trig [#permalink]  20 Jul 2008, 10:18
durgesh79 wrote:
you should never take diagram given in geomatry question as an accurate one. If i change the diagram to this, I'm sure no one will answer $$sqrt(3)$$

the 30-60-90 traingle mentioned by JasonC are actually these two traingles.

Attachment:
circle.JPG

aah ... I see what you are saying. the two triangles are only sharing the hypotenus. the bases are exchanged. Thnaks ...

PS:- GOD! when am I gonna stop doing these stupid mistakes.
Re: solve without trig   [#permalink] 20 Jul 2008, 10:18
Similar topics Replies Last post
Similar
Topics:
how can i solve this problem without assuming x ? 4 07 Aug 2014, 02:56
3 solve this ;) 3 30 Oct 2013, 11:50
1 Solving without plugging in numbers? 6 28 Feb 2010, 10:00
solve 1 23 Mar 2009, 20:50
solving this 3 03 Jun 2006, 00:34
Display posts from previous: Sort by