GMAT Prep XY Coordinate with Half Circle

dnkcdnguy wrote:This one took me awhile and I kind of guessed that it was "1".

I'm interested to see how others solve this quesiton.

Thanks

Draw vertical line from P to X axis. Call that point A.
Draw vertical line from Q to X axis. Call that point B.
PAO is a 30-60-90 triangle.
OP=2
OQ=2
<POA=30
<QOB=60
QOB is 30-60-90 triangle.
OB=1

the answer is D
if you drop the line from point Q the value is the square rout of 3

griscomtestprep wrote:

dnkcdnguy wrote:This one took me awhile and I kind of guessed that it was "1".

I'm interested to see how others solve this quesiton.

Thanks

Draw vertical line from P to X axis. Call that point A.
Draw vertical line from Q to X axis. Call that point B.
PAO is a 30-60-90 triangle.
OP=2
OQ=2
<POA=30
<QOB=60
QOB is 30-60-90 triangle.
OB=1

good approach..!! without solving.. we can answer the question

I did it in another approach

slope of the line OP = 1-0/-SQRT(3)-0 = - 1/SQRT(3)

slope of the line OQ, which is perpendicular to OP = sqrt(3) = t-0/s-0 = t/s
radius=op =
{(-sqrt(3))}^{2} + 1^2=4 =OQ= 4=
{s*sqrt(3)}^2+ s^2
--> s=1

<POA=30
<QOB=60
QOB is 30-60-90 triangle.

How did you figure out that they were 30 60 90 triangles? Couldn't they be 45-45-90 triangles as well?

For any right-angled triangle,
- if 3 internal angles are 30-60-90, the ratio of triangle's 3 sides is 1:sqrt3:2; and vice versa
- if 3 internal angles are 45-45-90, the ratio of triangle's 3 sides is 1:1:sqrt2; and vice versa

In the original question, if you draw a vertical like from P to axis X, and intersect X at point A, then triangle PAO is 30-60-90.

Since it's given that angle POQ is a right angle, if you draw a vertical like from Q to axis X and interset at point B, QBO is 30-60-90 triangle.

pakaskwa wrote:For any right-angled triangle,
- if 3 internal angles are 30-60-90, the ratio of triangle's 3 sides is 1:sqrt3:2; and vice versa
- if 3 internal angles are 45-45-90, the ratio of triangle's 3 sides is 1:1:sqrt2; and vice versa

In the original question, if you draw a vertical like from P to axis X, and intersect X at point A, then triangle PAO is 30-60-90.

Since it's given that angle POQ is a right angle, if you draw a vertical like from Q to axis X and interset at point B, QBO is 30-60-90 triangle.

You have not answered his question. Why can Q B O not be 45 is his question. The only facts you know according the problem is that the other two angles along the x axis sum to 90 and the two radii are each 2. How do these facts force QBO to be 60?.

How can you determine (or rather assume) that the triangle you are referring to, is a 30-60-90 trinagle?

According to me , the question does not provide enough data to support the arguement that whether the triangle(as referred earlier) will be 30-60-90 or 45-90-45.

So you need to make an assumption here.

From the figure, one can see that PO and QO are equially distanced from Y axis as the Y axis bisects the 90 degree angle symbol.

So, I assume that s = negative of (-sqrt(3)), i.e. sqrt(3).

Therefore, D seems to be the right answer.

Whats the OA?

The small square in the angle means right angle… meaning 90 degrees.

So only use logic.

Imaging a point (0,1) the radius is 1 (circle)
Add the fact that is a 90 degrees figure

The other point will be the opposite (1,0)

In this case the initial point is – sqr3 , 1 so the opposite is 1, sqr3

Answer 1

Explanation for why it isnt 45 45 90 triangle
I am not sure, if it is ok to give an explanation wrt trigonometry as to how much it is used in GMAT land. But it is the simplest thing to put here.

In triangle PAO, where A is the point on x-axis directly below P, and just for the sake of finding the angle, let's forget that we are in second quadrant. So let's take point P as {root(3), 1}
OP = 2
PA = 1
AO = root(3)

Sin (x) = PA/OP = 1/2
We know that Sin(30) is 1/2. That implies Angle POA is 30 degrees.

Hope this helps.