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A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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07 Jun 2012, 00:16

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A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many feet above the ground does the ladder reach?

Notice that trigonometry is not tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.

sarb wrote:

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many meet above the ground does the ladder reach?

A. 35 B. 42 C. 35 root 3 D. 7 + 35 root 3 E. 7 + 42 root 3

Look at the diagram below:

Triangle ABC is a 30°-60°-90° triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (AC) corresponds with \(1\), the leg opposite 60° (BC) corresponds with \(\sqrt{3}\) and the hypotenuse AB corresponds with 2. So, \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\) --> \(\frac{BC}{70}=\frac{\sqrt{3}}{2}\) --> \(BC=35\sqrt{3}\).

Hence the leader reaches \(7+35\sqrt{3}\) above the ground.

Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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07 Jun 2012, 18:21

Bunuel wrote:

Notice that trigonometry is not tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.

sarb wrote:

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many meet above the ground does the ladder reach?

A. 35 B. 42 C. 35 root 3 D. 7 + 35 root 3 E. 7 + 42 root 3

Look at the diagram below:

Attachment:

Ladder.png

Triangle ABC is a 30°-60°-90° triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (AC) corresponds with \(1\), the leg opposite 60° (BC) corresponds with \(\sqrt{3}\) and the hypotenuse AC corresponds with 2. So, \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\) --> \(\frac{BC}{70}=\frac{\sqrt{3}}{2}\) --> \(BC=35\sqrt{3}\).

Hence the leader reaches \(7+35\sqrt{3}\) above the ground.

Concentration: Entrepreneurship, International Business

GMAT 1: 730 Q50 V39

GPA: 3.2

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Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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21 Feb 2013, 09:38

Since the questions says elevated at an angle of 60 deg, the base angle is 60 deg. By default, elevation is from baseline and since the ladder is placed parallel to x axis, the elevation is also from x axis.

Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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29 Dec 2013, 10:00

I understand the ratio x:x(sqrt)3:2x and how to get 7+35sqrt3, but how is this not a 3-4-5 right triangle with lengths of 42-56-70? Thank you for the added explanation.

I understand the ratio x:x(sqrt)3:2x and how to get 7+35sqrt3, but how is this not a 3-4-5 right triangle with lengths of 42-56-70? Thank you for the added explanation.

If it's 42-56-70, what is x then? Also, we get that BC, the smallest side, is \(35\sqrt{3}\) not 42.

Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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20 Sep 2014, 00:55

sarb wrote:

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many feet above the ground does the ladder reach?

A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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24 Mar 2015, 22:43

Hi All,

Another approach to solve this question is if you know the values of sin/cos angles. In this case sin 60 = Perpendicular/Hypotenuse i.e. sin 60 = BC/AB. As sin 60 =\(\sqrt{3}/2\), we get \(\sqrt{3}/2\) = BC/70, thus BC = 35\(\sqrt{3}\)

From that you can get total height from the base 7 + 35\(\sqrt{3}\) as the final answer. Hope that helps.

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Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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02 Nov 2015, 08:45

Bunuel wrote:

Notice that trigonometry is not tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.

sarb wrote:

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many meet above the ground does the ladder reach?

A. 35 B. 42 C. 35 root 3 D. 7 + 35 root 3 E. 7 + 42 root 3

Look at the diagram below:

Attachment:

Ladder.png

Triangle ABC is a 30°-60°-90° triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (AC) corresponds with \(1\), the leg opposite 60° (BC) corresponds with \(\sqrt{3}\) and the hypotenuse AC corresponds with 2. So, \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\) --> \(\frac{BC}{70}=\frac{\sqrt{3}}{2}\) --> \(BC=35\sqrt{3}\).

Hence the leader reaches \(7+35\sqrt{3}\) above the ground.

I understand this but I drew it sideways and got the answer of (B) 42

I'm not sure how you determined which side is the truck and which is the wall. Couldn't it have been a very long truck and a short wall that the ladder was against?

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many feet above the ground does the ladder reach?

We are given that the ladder of a fire truck is elevated to an angle 60 degrees above the ground and that the ladder has a length of 70 feet. We are also given that the ladder is 7 feet above the ground. The best thing to do in this situation is to draw a diagram.

Notice that the resulting triangle in the sketch is a 30-60-90 right triangle. Based on the given info, we don’t know that the ladder is leaning against a building whose side is perpendicular to the ground. The ratio of the sides of a 30-60-90 right triangle is x : x√3 : 2x. We see that the hypotenuse length of 70 feet is equal to the "2x" from the 30-60-90 ratio. Thus, we can set up an equation and solve for x.

70 = 2x

x = 35

Because x = 35, we know that the side opposite the 60-degree angle or, in this case, the height of the ladder, is 35√3. The height of the ladder is 35√3 feet and the base of the ladder is 7 feet above the ground; thus, we know that the ladder reaches a total height above the ground of 35√3 + 7 feet.

Answer D
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A ladder of a fire truck is elevated to an angle of 60° and [#permalink]

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27 Nov 2017, 12:41

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Bunuel wrote:

Notice that trigonometry is not tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.

sarb wrote:

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many meet above the ground does the ladder reach?

A. 35 B. 42 C. 35 root 3 D. 7 + 35 root 3 E. 7 + 42 root 3

Look at the diagram below:

Triangle ABC is a 30°-60°-90° triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (AC) corresponds with \(1\), the leg opposite 60° (BC) corresponds with \(\sqrt{3}\) and the hypotenuse AC corresponds with 2. So, \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\) --> \(\frac{BC}{70}=\frac{\sqrt{3}}{2}\) --> \(BC=35\sqrt{3}\).

Hence the leader reaches \(7+35\sqrt{3}\) above the ground.

Notice that trigonometry is not tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.

sarb wrote:

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many meet above the ground does the ladder reach?

A. 35 B. 42 C. 35 root 3 D. 7 + 35 root 3 E. 7 + 42 root 3

Look at the diagram below:

Triangle ABC is a 30°-60°-90° triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (AC) corresponds with \(1\), the leg opposite 60° (BC) corresponds with \(\sqrt{3}\) and the hypotenuse AC corresponds with 2. So, \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\) --> \(\frac{BC}{70}=\frac{\sqrt{3}}{2}\) --> \(BC=35\sqrt{3}\).

Hence the leader reaches \(7+35\sqrt{3}\) above the ground.