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Re: In the figure above, V represents an observation point at on [#permalink]
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madn800 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?
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atl12688 wrote:
madn800 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]
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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Re: In the figure above, V represents an observation point at on [#permalink]
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See Image for plugging in strategy
Attachments

Observatory.jpg
Observatory.jpg [ 44.41 KiB | Viewed 27260 times ]

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


We are being asked to determine the length of RS. To determine this length we need to know the length from point R to the right angle in the given figure. If we label the point at the right angle as T, we see that we need to determine the length of TR.

If we draw a line segment connecting V and R, we will see that VR, VT and TR create a right triangle. Furthermore, we are told in the question stem that VR (the hypotenuse) is 10, and that one of the sides, VT, is 5, so we now plug these values into the Pythagorean Theorem.

TR^2 + VT^2 = VR^2

TR ^2 + 5^2 = 10^2

TR ^2 + 25 = 100

TR ^2 = 75

TR = √75

TR = √25 x √3

TR = 5√3

So TR is 5√3. We subtract this from the total length TS, which is 10, to determine the length from R to S:

10 - 5√3

Answer is A.
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Re: In the figure above, V represents an observation point at on [#permalink]
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Walkabout wrote:
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) \(2 \frac{1}{2}\)
(E) \(4\)


Given: VR = 10 feet
Let's add a line from V to R to create the following right triangle:


Apply the Pythagorean theorem to get: \(x^2 + 5^2 = 10^2\)
Simplify: \(x^2 + 25 = 100\)
Subtract \(25\) from both sides of the equation: \(x^2 = 75\)
Solve to get: \(x = \sqrt{75}\)

To simplify \(\sqrt{75}\), we'll use the fact that \(\sqrt{(x)(y)}=(\sqrt{x})(\sqrt{y})\)

So, \(\sqrt{75}=\sqrt{(25)(3)}\)
\(=(\sqrt{25})(\sqrt{3})\)
\(=5\sqrt{3}\)

Aside: Another way to find the value of x is to first recognize that, since the hypotenuse is TWICE the length of one leg, we must have a 30-60-90 right triangle on our hands, in which case x=5√3

Our diagram now looks like this:


What is the distance RS?
From the diagram, we can see that \(=5\sqrt{3}+RS = 10\), which means \(RS = 10-5\sqrt{3}\)

Answer: A
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