Gmat Prep Geometry

[steinbock]

Stuck on this one....does anyone know how to solve this?

[GMATDavid]

The triangle is a special right triangle (30-60-90). The ratios of the sides are x, x-root3 and 2x, respectively. Because the y value is 1, the little angle adjacent to the right angle is 30-degrees. That means the angle on the other side of the right angle is 60 degrees, forming another 30-60-90 triangle. The question asks for the side opposite the 30-degree angle of that triangle, so the answer is 1. The diagram is not to scale -- you just have to apply the rules and ignore how it looks.

[GMATDavid]

The triangle is a special right triangle (30-60-90). The ratios of the sides are x, x-root3 and 2x, respectively. Because the y value is 1, the little angle adjacent to the right angle is 30-degrees. That means the angle on the other side of the right angle is 60 degrees, forming another 30-60-90 triangle. The question asks for the side opposite the 30-degree angle of that triangle, so the answer is 1. The diagram is not to scale -- you just have to apply the rules and ignore how it looks.

[GMATDavid]

The triangle is a special right triangle (30-60-90). The ratios of the sides are x, x-root3 and 2x, respectively. Because the y value is 1, the little angle adjacent to the right angle is 30-degrees. That means the angle on the other side of the right angle is 60 degrees, forming another 30-60-90 triangle. The question asks for the side opposite the 30-degree angle of that triangle, so the answer is 1. The diagram is not to scale -- you just have to apply the rules and ignore how it looks.

[rs2010]

Since one point is given so you can calculate radius of the circle which will
2. After this calculate the length of arc between these points which will be equal to 2*pi*r/4 ( since angle is 90 degrees).

This will give you values of s and t.

[steinbock]

Thanks to both of you for your explanations. Still trying to understand though....

[rs2010]

Since O is center so distance would be equal to radius from any point on the circle.
Second is length of arc e.g.

If you have 360 degree angle then its called the circuference with 2*pi*r and if you have any angle then you will do angle/360degrees.


Let me know if this helps.

[lion147]

Thanks to both of you for your explanations. Still trying to understand though....

Try drawing it to scale on paper.

Point P forms a 30-60-90 triangle with the X axis, the base is sroot(3), and the height is 1. We know the ratios of the sides (1:sroot(3):2) so the longest side, which is also the circle's radius, is 2x1 = 2. Also the angle formed by the line to P and the X axis is 30 degrees.

Now, you can calculate the size of the angle formed by the line to point Q and the X axis, it's 180-90-30 = 60 degrees.

So there's another 30-60-90 triangle, the base is 's', the height is 't' and the longest side is 2. Again we know the ratios, and we have the angles so:

s = 2/2 = 1
t = (2/2) x sroot(3) = sroot(3).

[onesome]

Can you please explain how you are identifying that this triangle is a special triangle = 30-60-90 triangle?
I can see 90 but Im not sure how you conclude 30-60 for other 2.

[Stuart@KaplanGMAT]

Here's another thread on this question:

https://www.beatthegmat.com/please-help- ... t7941.html

[lion147]

Can you please explain how you are identifying that this triangle is a special triangle = 30-60-90 triangle?
I can see 90 but Im not sure how you conclude 30-60 for other 2.

The first triangle on the left has the X axis as it's base with a length of sroot(3), the height is 1 and there's a right angle in between; this must be a 30-60-90 triangle.

The same triangle is on the right, but rotated; the radius (2) is the longest side, we also have two angles - 30 and 90, so this also must be a 30-60-90 triangle.

Also, you don't need to use the 30-60-90 rule, you can use pythag once you know one angle is 90 degrees, but it would be better to know the shortcut as a time saver.

[GMATDavid]

Yes, the 30-60-90 rule is very convenient to know as the GMAT test writers like to use it.

All 30-60-90 triangles have sides with the same ratios. x, x-root3 and 2x. So you could have 1, root3, 2 as here, or 3, 3-root3, 6 in a bigger triangle.

Every diagonal line in the coordinate plane is the hyptenuse of a right triangle. Here, using the x and y values (the two sides) you can recognize the 30-60-90 ratio.

[Ian Stewart]

I could suggest a different approach here. Notice that OP and OQ are perpendicular lines, and are the same length. We can use the simple fact that they are perpendicular to get the answer directly here, with no calculations.

To review the theory (and I'll use concrete numbers to make things easier): if 2/3 is the slope of a line L, we know that -3/2 is the slope of a perpendicular line M. Now, what does the slope '2/3' mean? It means you need to go right 3 units in order that L will rise by 2 units. That's the definition of the slope. On the perpendicular line M, with slope -3/2, you would need to go right 2 units in order for the line to 'rise' -3 units (i.e. fall by 3 units). The vertical and horizontal changes get 'reversed' on the perpendicular line, and one of the changes becomes negative. Thus, if (0,0) is a point on both L and M, then we know that (3, 2) will be a point on L, and (2, -3) will be a point on the perpendicular line M. Notice that each of these points is the same distance from (0,0).

The relationship between perpendicular lines becomes much clearer if you draw a few examples on the co-ordinate plane. Anyway, if you understand this relationship well, the question can be solved almost immediately:

On OP, if we go right by sqrt(3) units, the line falls by 1 unit. On the perpendicular line OQ, if we go right by 1 unit, the line must rise by sqrt(3) units, since OQ is perpendicular to OP. Thus, point Q must be (1, sqrt(3)).

[Ian Stewart]

Aha, I just followed the link Stuart posted above, and a while ago he suggested the same solution as I just did, with a very clear explanation as well- nice one!