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A ladder of a fire truck is elevated to an angle of 60° and [#permalink]
07 Jun 2012, 00:16

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Question Stats:

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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?

Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]
07 Jun 2012, 10:50

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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 [ 5.44 KiB | Viewed 14201 times ]

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.

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

Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]
21 Feb 2013, 09:31

which one of 2 angles is 60?

the question is not clear. _________________

if anyone in this forum is living in Bradford, Leeds or nearby areas, pls, email to me, thanghnvn@gmail.com. I have a small thing to ask you. Thank you

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

Re: A ladder of a fire truck is elevated to an angle of 60° and [#permalink]
29 Dec 2013, 10:16

Expert's post

bgourlay13 wrote:

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]
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]
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.

Attachments

File comment: Trigonometry angles

trigonometry-angles4.jpg [ 7.53 KiB | Viewed 485 times ]

gmatclubot

A ladder of a fire truck is elevated to an angle of 60° and
[#permalink]
24 Mar 2015, 22:43

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