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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
The angle made by the line \sqrt{3}x+y+6=0, with y-axis is
1. 60 degrees 2. 30 degrees 3. 45 degrees 4. 120 degrees 5. None of the above
We have the line \(y=-\sqrt{3}*x-6\).
Attachment:
untitled.PNG [ 14.87 KiB | Viewed 4349 times ]
Find the x-intercept of the line (x-intercept of a line the value of x for y=0, the point where the line crosses X-axis): y=0 --> \(x=-2\sqrt{3}\);
Find the y-intercept of the line (y-intercept of a line the value of y for x=0, the point where the line crosses Y-axis): x=0 --> \(y=-6\);
In the right triangle created by the given line and the axis the ratio of lengths of the legs is \(2\sqrt{3}:6=1:\sqrt{3}\) so we have 30°, 60°, and 90° right triangle (in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).
Now, the angle at y-axis is opposite the smallest side \(2\sqrt{3}\) so it equals to 30°.
Answer: B.
Or you can notice that as the slope, which is th ratio of the rise over run is equal to \(-\sqrt{3}\) then the ratio of the length of the leg at y-axis over the length of the leg at x-axis will be \(\sqrt{3}\), which again tells us that we have 30°, 60°, and 90° right triangle.
that was smart. I tried to use m = tan(degree), Is this wrong. If no can you please explain how to compute using this.
Every GMAT geometry question can be solved without trigonometry, so you don't need trigonometry for GMAT.
But yes, \(slope=m=tan(\alpha)\), where \(\alpha\) is the angle line makes with positive x-axis, so as \(m=-\sqrt{3}=tan(\alpha)\) --> \(\alpha=120^{\circ}\) --> angle at y-axis will be 120-90=30. _________________
This question is incomplete, for answer to be 30 degrees, question should ask the smaller angle between line and y-axis. Otherwise answer can be either 150 degrees or 30 degrees. _________________
The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!
This question is incomplete, for answer to be 30 degrees, question should ask the smaller angle between line and y-axis. Otherwise answer can be either 150 degrees or 30 degrees.
Though I see your point, still only one answer from 150 and 30 is listed among answer choices, so we can safely choose 30 as correct one. _________________
The angle made by the line \sqrt{3}x+y+6=0, with y-axis is
1. 60 degrees 2. 30 degrees 3. 45 degrees 4. 120 degrees 5. None of the above
We have the line \(y=-\sqrt{3}*x-6\).
Attachment:
untitled.PNG
Find the x-intercept of the line (x-intercept of a line the value of x for y=0, the point where the line crosses X-axis): y=0 --> \(x=-2\sqrt{3}\);
Find the y-intercept of the line (y-intercept of a line the value of y for x=0, the point where the line crosses Y-axis): x=0 --> \(y=-6\);
In the right triangle created by the given line and the axis the ratio of lengths of the legs is \(2\sqrt{3}:6=1:\sqrt{3}\) so we have 30°, 60°, and 90° right triangle (in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).
Now, the angle at y-axis is opposite the smallest side \(2\sqrt{3}\) so it equals to 30°.
Answer: B.
Or you can notice that as the slope, which is th ratio of the rise over run is equal to \(-\sqrt{3}\) then the ratio of the length of the leg at y-axis over the length of the leg at x-axis will be \(\sqrt{3}\), which again tells us that we have 30°, 60°, and 90° right triangle.
Can you pls explain how did draw the graph... How did you take the points of x and y..
\(y=mx+b\) is called point-intercept form of equation of a line (in our case it's \(y=-\sqrt{3}*x-6\)), where \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\). To draw a graph of a line you need two points (any two distinct points (x,y) which satisfy the equation of a line) and then you just draw a line passing these points. Usually X and Y intercepts are best choices for the points.
X-intercept is the point where a line (a graph) crosses the x-axis. So it's the point on x-axis, any point on x-axis has y-coordinate equal to zero, which means that X-intercept is the point \((x,0)\) - the value of \(x\) when \(y=0\): \(y=mx+b\) --> \(0=mx+b\) --> \(x=-\frac{b}{m}\). So X-intercept of a line \(y=mx+b\) is \(x=-\frac{b}{m}\);
Y-intercept is the point where a line (a graph) crosses the y-axis. So it's the point on y-axis, any point on y-axis has x-coordinate equal to zero, which means that Y-intercept is the point \((0,y)\) - the value of \(y\) when \(x=0\): \(y=mx+b\) --> \(y=m*0+b\) --> \(y=b\). So Y-intercept of a line \(y=mx+b\) is \(y=b\).
In our case for the line given by \(y=-\sqrt{3}*x-6\): X-intercept will be \((-2\sqrt{3}, \ 0)\), as x-intercept of a line is the value of x for y=0, the point where the line crosses X-axis: y=0 --> \(x=-2\sqrt{3}\);
Y-intercept will be \((0, \ -6)\), as y-intercept of a line is the value of y for x=0, the point where the line crosses Y-axis: x=0 --> \(y=-6\).
Can you pls explain how did draw the graph... How did you take the points of x and y..
\(y=mx+b\) is called point-intercept form of equation of a line (in our case it's \(y=-\sqrt{3}*x-6\)), where \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\). To draw a graph of a line you need two points (any two distinct points (x,y) which satisfy the equation of a line) and then you just draw a line passing these points. Usually X and Y intercepts are best choices for the points.
X-intercept is the point where a line (a graph) crosses the x-axis. So it's the point on x-axis, any point on x-axis has y-coordinate equal to zero, which means that X-intercept is the point \((x,0)\) - the value of \(x\) when \(y=0\): \(y=mx+b\) --> \(0=mx+b\) --> \(x=-\frac{b}{m}\). So X-intercept of a line \(y=mx+b\) is \(x=-\frac{b}{m}\);
Y-intercept is the point where a line (a graph) crosses the y-axis. So it's the point on y-axis, any point on y-axis has x-coordinate equal to zero, which means that Y-intercept is the point \((0,y)\) - the value of \(y\) when \(x=0\): \(y=mx+b\) --> \(y=m*0+b\) --> \(y=b\). So Y-intercept of a line \(y=mx+b\) is \(y=b\).
In our case for the line given by \(y=-\sqrt{3}*x-6\): X-intercept will be \((-2\sqrt{3}, \ 0)\), as x-intercept of a line is the value of x for y=0, the point where the line crosses X-axis: y=0 --> \(x=-2\sqrt{3}\);
Y-intercept will be \((0, \ -6)\), as y-intercept of a line is the value of y for x=0, the point where the line crosses Y-axis: x=0 --> \(y=-6\).
But why the tan(@) formula works on this. As Y axis has a slope = infinity. So this formula doesn't work and i had chosen last option which says "None of the above". I know this option may not be there in GMAT but why can't we apply a particular formula to reach the same result.
Tan(@) = |(m1 - m2)/(1+m1*m2)|
Where m1 and m2 are the slopes of two lines and "@" is angle between these line.
The angle made by the line \sqrt{3}x+y+6=0, with y-axis is
1. 60 degrees 2. 30 degrees 3. 45 degrees 4. 120 degrees 5. None of the above
Or you can notice that as the slope, which is th ratio of the rise over run is equal to \(-\sqrt{3}\) then the ratio of the length of the leg at y-axis over the length of the leg at x-axis will be \(\sqrt{3}\), which again tells us that we have 30°, 60°, and 90° right triangle.
Hope it's clear.
Hi Bunuel , Sorry to dig in an old post . I am unable to follow your second approach properly (i.e how did you find the ratio length of y to length of x as sqrt3) . Could you please elaborate it more .
The explanation of this question is fantastic and one can quickly solve these kind of questions after understanding it. The only concept that i am not able to understand is how \(2\sqrt{3}:6\) turn into \(1:\sqrt{3}\)
I will really appreciate if anyone can help me in understanding this concept.
Convert the given equation in this form to get y=-rt3x -6.
Since quesn deals in slope you need not calculate intercepts, clearly slope is negative and is tan(rt30 which is 60 degree. It lies in 3rd quadrant Since slope is defined as what angle a line makes with x -axis, the angle with y axis is 90-60 = 30.
If you want to know about intercepts, the above eqn cut x axis when y=0 rt? So when y=0, x = -6/rt3 = 3* 2 / rt3 = -2rt3. Also when x=0, y=-6.
Re: The angle made by the line root{3}x+y+6=0, with y-axis is [#permalink]
12 Oct 2013, 06:01
"The explanation of this question is fantastic and one can quickly solve these kind of questions after understanding it. The only concept that i am not able to understand is how 2\(\sqrt{3}\):6 turn into 1:\(\sqrt{3}\)"
here goes first cancel with 2 on either side of the ratio \(\sqrt{3}\) : 3 on the right 3 can be written as \(\sqrt{3}\)*\(\sqrt{3}\) \(\sqrt{3}\) : \(\sqrt{3}\)* \(\sqrt{3}\) cancel \(\sqrt{3}\) you get 1: \(\sqrt{3}\)
Re: The angle made by the line root{3}x+y+6=0, with y-axis is [#permalink]
12 Nov 2013, 03:07
Another idea: Observe buneul's diagram. The coordinates on the -ve x-axis and -ve y-axis are (-2\(\sqrt{3}\),0) and (0,-6). Now find the distance between them using the formula \(\sqrt{x^2+y^2}\). We get 4\(\sqrt{3}\)=2*2\(\sqrt{3}\). Now we know that the ratio of sides for a 30-60-90 triangle is x:\(\sqrt{3}\)x:2x. From distance calculation above we can see that the following ratio holds perfectly good 30:60:90=2\(\sqrt{3}\): \(\sqrt{3}\)*(2\(\sqrt{3}\)): 2*(2\(\sqrt{3}\)). Comparing we can say that side opposite to 2\(\sqrt{3}\) measures 30 degrees.
Re: The angle made by the line root{3}x+y+6=0, with y-axis is [#permalink]
29 Jul 2015, 20:30
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...