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The top of a 15 metre high tower makes an angle of elevation of 60° wi

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The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 01 Jul 2020, 22:02
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The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

A. 5 metres
B. 8 metres
C. 10 metres
D. 12 metres
E. 13 metres

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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post Updated on: 01 Jul 2020, 22:27
Let AB be the tower and CD be the electric pole.
\(∠ACB=60∘ , ∠EDB=30∘ , AB=15\)
\(Let BE=AB−AE=15−h\)
\(\frac{AB}{AC}=tan 60∘= \sqrt{3}\)
=>\(AC=\frac{AB}{\sqrt{3}}\)
=>\(AC=\frac{15}{\sqrt{3}}\)
Also,
\(\frac{BE}{DE}=tan 30∘=\frac{1}{\sqrt{3}}\)
=>\(DE=BE\sqrt{3}=\sqrt{3}(15−h)\)

AC = DE
=> \(\frac{15}{\sqrt{3}} = \sqrt{3}(15−h)\)
=> \(15 = 45 - 3h\)
=> \(3h = 30\)
=> \(h = 10\)

Hence, OA is (C).
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Originally posted by v12345 on 01 Jul 2020, 22:26.
Last edited by v12345 on 01 Jul 2020, 22:27, edited 1 time in total.
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 01 Jul 2020, 22:27
Explanation:
Check Figure below for better understanding:
∠ACB = 60 , ∠EDB = 30 , AB=15 , AC=DE
BE = AB − AE =15−h
AB/AC=tan60 = √3
AC=15/√3

BE/DE= tan30 = 1/√3
DE=√3*(15−h)

As AC = DE
15/√3 = √3*(15−h)
15 = 45 - 3h
h = 30/3
h = 10

IMO-C
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 01 Jul 2020, 22:28
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 01 Jul 2020, 22:35
We need to apply the concept of 30-60-90 triangle.
We are given that angle of elevation with the bottom of the electric pole is 60° so \(\angle ADC = 60°\)
In \(\triangle ACD, \angle ACD = 90°\)
So, \(\angle CAD = 30°\)

Now, we can apply the ratio of 30-60-90.
\(1:\sqrt{3}:2\)

So, the side opposite \(\angle CAD = CD = \frac{AC}{\sqrt{3}} = \frac{15}{\sqrt{3}}\)

We are given that angle of elevation with the top of the electric pole is 30° so \(\angle AEB = 30°\)
In \(\triangle ABE, \angle ABE = 90°\)
So, \(\angle BAE = 60°\)

Now, we can apply the ratio of 30-60-90.
\(1:\sqrt{3}:2\)

We know that \(BE = CD = \frac{15}{\sqrt{3}}\)

So, the side opposite \(\angle AEB = AB = \frac{BE}{\sqrt{3}} = \frac{15}{(\sqrt{3} \times \sqrt{3})} = \frac{15}{3} = 5\)

Height of the electric pole = \(ED = BC = AC - AB = 15 - 5 = 10\)

OA, C
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 02 Jul 2020, 01:04
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This is a question which tests your knowledge of the 30-60-90 right-angled triangle. As such, let us do a quick recap on the properties of a 30-60-90 right triangle.
The figure given below shows a standard 30-60-90 right triangle:

Attachment:
02nd July 2020 - Reply 2 - 1.jpg
02nd July 2020 - Reply 2 - 1.jpg [ 37.04 KiB | Viewed 426 times ]


From the figure, we gather that, if the side opposite to the 30-degree angle is ‘x’ units, then the side opposite to the 60-degree angle will be x√3 units and the hypotenuse will be 2x units.
Alternatively, if the side opposite to the 60-degree angle is ‘x’ units, then the side opposite to the 30-degree angle will be \(\frac{x}{√3}\) units.

We can draw another diagram to represent the situation described in the question, to look like the one in the image shown below:

Attachment:
02nd July 2020 - Reply 2 - 2.jpg
02nd July 2020 - Reply 2 - 2.jpg [ 56.4 KiB | Viewed 425 times ]


From the figure, the height of the electric pole is evidently 10 metres.
The correct answer option is C.

Hope that helps!
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 02 Jul 2020, 01:32
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IMO C

Ref attached Image.
AB= Tower = 15 m
CD= Pole
∠ ACE=30, ∠ADB=60

In △ ADB
tan 60 = AB/DB
DB = AB/tan 60

In △ ACE
tan 30 = AE/CE
AE= CE tan 30 = DB tan 30= AB/tan 60 x tan 30 = 15 x 1/√3 x 1/√3 = 5

Therefore, height of pole= CD= 15-5 = 10 m


C. 10 metres
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 02 Jul 2020, 04:35
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AB is the tower and CD is the pole.

AB = 15

AB : BC : AC = root(3) : 1 : 2

=> BC = 15/root(3) = DE

Similarly, AE : DE : AD = 1 : root(3) : 2

=> AE = 15/root(3) * 1/root(3) = 15/3 = 5

CD = BE = AB - AE = 15 - 5 = 10

Ans: C
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 02 Jul 2020, 11:55
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Ans - C

Draw the diagram. Use two triangles and find distance btw pole and tower, height of pole

Let x be height of pole and y be distance btw pole and tower
Tan30 = 15-x / y
Tan60 = 15/y
Y= 15/ _/3
Put this in first equation

You get x = 10

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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 02 Jul 2020, 20:28
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The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

A. 5 metres
B. 8 metres
C. 10 metres
D. 12 metres
E. 13 metres

angle with the bottom of pole is 60 degree
so the triangle is 30 60 90
15 m is opp of 60,
so distance between pole and tower is 5 * root3

angle with top of pole is 30 degree
so 30 60 90
5 * root3 is opp of 60
so the extra height, opp of 30 is 5

therefore, the height of pole is 15-5=10

Ans C
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 02 Jul 2020, 21:15
1
given values are:
Their are electronic pole & high tower.
Height of High tower is 15m
Top of high tower makes 60deg angle with bottom of electronic pole:
it means triangle formed by high pole & bottom of electronic pole will have angle 30, 60, 90 with height 15m.
if internal angles are in 30, 60, 90, corresponding sides will be in x, (3)^1/2x, & 2x.
With that base (distance b/w high pole & electronic tower) will come as 15/(3)^1/2.

Now top of electronic pole makes 30 deg angle with high tower. it means again the triangle made will be 30, 60, 90 with hypotenuse joining both top with 30 deg angle. using the same formula of 30, 60, 90 triangle with base (distance b/w high pole & electronic tower) already identified as 15/(3)^1/2.
perpendicular will come as 5 (portion of high pole in triangle)

since high pole is 15m, so height of the electronic pole will be 10m (C)
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi  [#permalink]

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New post 03 Jul 2020, 00:55
We get similar triangles of 30-60-90°.

Answer C = 10m
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Re: The top of a 15 metre high tower makes an angle of elevation of 60° wi   [#permalink] 03 Jul 2020, 00:55

The top of a 15 metre high tower makes an angle of elevation of 60° wi

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