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# In the figure above, V represents an observation point at on

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In the figure above, V represents an observation point at on [#permalink]

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19 Dec 2012, 06:05
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In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object?

(A) $$10-5\sqrt{3}$$
(B) $$10-5\sqrt{2}$$
(C) 2
(D) 2 1/2
(E) 4
[Reveal] Spoiler: OA
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Re: In the figure above, V represents an observation point at on [#permalink]

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19 Dec 2012, 06:09
Expert's post

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object?

(A) $$10-5\sqrt{3}$$
(B) $$10-5\sqrt{2}$$
(C) 2
(D) 2 1/2
(E) 4

Attachment:

observation2.png [ 12.99 KiB | Viewed 7644 times ]

$$PR=\sqrt{VR^2-VP^2}=\sqrt{10^2-5^2}=5\sqrt{3}$$;

Thus, $$RS=PS-PR=10-5\sqrt{3}$$.

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Re: In the figure above, V represents an observation point at on [#permalink]

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17 Nov 2013, 20:38
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No need of using Pythagorean theorem. Observe triangle VPR. VP:VR=5:10=1:2. Angle opposite VR is 90. When does this happen? It happens only when VPR is a 30-60-90 triangle. So, VP:VR:PR=5:10:5$$\sqrt{3}$$.So, answer is 10-5$$\sqrt{3}$$
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Re: In the figure above, V represents an observation point at on [#permalink]

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16 Apr 2014, 11:04
No need of using Pythagorean theorem. Observe triangle VPR. VP:VR=5:10=1:2. Angle opposite VR is 90. When does this happen? It happens only when VPR is a 30-60-90 triangle. So, VP:VR:PR=5:10:5$$\sqrt{3}$$.So, answer is 10-5$$\sqrt{3}$$

How can you determine this is a 30-60-90 and not a 45-45-90?
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Re: In the figure above, V represents an observation point at on [#permalink]

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18 Apr 2014, 11:05
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atl12688 wrote:
No need of using Pythagorean theorem. Observe triangle VPR. VP:VR=5:10=1:2. Angle opposite VR is 90. When does this happen? It happens only when VPR is a 30-60-90 triangle. So, VP:VR:PR=5:10:5$$\sqrt{3}$$.So, answer is 10-5$$\sqrt{3}$$

How can you determine this is a 30-60-90 and not a 45-45-90?

In right triangle VPR the ratio of one side (VP) to hypotenuse (VR) is 1:2. This only happens for 30-60-90 right triangle.

MUST KNOW FOR THE GMAT:
• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: 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°).

• A right triangle where the angles are 45°, 45°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio $$1 : 1 : \sqrt{2}$$. With the $$\sqrt{2}$$ being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.

For more check Triangles chapter of our Math Book: math-triangles-87197.html

Hope it helps.
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Re: In the figure above, V represents an observation point at on [#permalink]

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18 Apr 2014, 11:30
Yes thank you Bunuel. I was missing the fact that VR was 10 which gave the Leg:hypotenuse the 1:2 ratio. Greatly appreciated!!
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In the figure above, V represents an observation point at on [#permalink]

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10 Jul 2014, 09:49
See Image for plugging in strategy
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Re: In the figure above, V represents an observation point at on [#permalink]

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27 Feb 2016, 11:51
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Re: In the figure above, V represents an observation point at on [#permalink]

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03 Mar 2016, 02:32
Attachment:
observation.png
In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object?

(A) $$10-5\sqrt{3}$$
(B) $$10-5\sqrt{2}$$
(C) 2
(D) 2 1/2
(E) 4

I think this is the fastest route to the answer.
Knowing the popular right-triangle and applying that knowledge when u meet a question of this sort rewards you massively.
2:1:√3 for 30 60 90
1: 1: √2 for 40 40 90.

And the fastest means of telling is finding the ratio of the sides regardless of if it's a 15360000: 7680000: x that you saw.
Re: In the figure above, V represents an observation point at on   [#permalink] 03 Mar 2016, 02:32
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