Hi, i ask this the other day and if someone can solve it with answer and explains, will fan and medal: When a beam makes an angle of 40 degrees with the ground, the top of the beam is 40 feet above the ground. There a telephone wires near by and the worker worried that the beam may hit the wires. When the beam makes a angle of 60 degrees with the ground, the wires are 2 feet above the beam. Will the beam hit the ground if the crew continues to raise it?

- anonymous

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- schrodinger

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- anonymous

@Nnesha

- Nnesha

looks like physics q lel :P
well i have to go
wait for someone or i'll llook at 't later !

- anonymous

okay :)

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## More answers

- radar

|dw:1439431270274:dw| I believe the question should be does the pole hit the WIRE as they continue to raise it, it will hit the ground if they drop it.!

- anonymous

@radar: i'm sorry, it suppose to be if it hit the wires or not.

- mathstudent55

|dw:1439437785859:dw|

- mathstudent55

Let's look at the beam at 40 deg to the ground.

- mathstudent55

|dw:1439437990141:dw|

- mathstudent55

\(\sin 40^o = \dfrac{40~ft}{L}\)
\(L = \dfrac{40~ft}{\sin 40^o} \approx 62.23 ~ ft\)

- mathstudent55

Now we know the length of the beam.
Now we look at the situation when the beam is at 60 degrees from the ground.

- mathstudent55

|dw:1439438459934:dw|

- mathstudent55

Notice I just corrected the length of the beam. I just noticed my calculator was set to radians and the length of the beam was incorrect. The length of the beam is 62.23 ft. This number is correct now.

- mathstudent55

Now we find h, the height of the beam when it is at a 60-deg angle with the ground.

- mathstudent55

\(\sin 60^o = \dfrac{h}{62.23~ft} \)
\(h = (62.23~ft)\sin 60^o \approx 53.89~ft \)

- mathstudent55

Since the wires are 2 ft higher than the top of the beam when the beam is at 60 deg with the ground, the wires are at
55.89 ft above the ground.

- mathstudent55

Since the beam is 62.23 ft long, and the wires are only 55.89 ft above ground, if the crew continues to raise the angle of the beam, the beam WILL HIT the wires.

- mathstudent55

At what angle of the beam will it hit the wires?
|dw:1439438842464:dw|

- anonymous

Beware the OS Code of Conduct!

- mathstudent55

\(\sin x = \dfrac{55.89}{62.23} \)
\(\sin^{-1} \dfrac{55.89}{62.23} = 63.91^o\)
At 63.91 deg, the beam will hit the wires.

- mathstudent55

@Leong Do you understand?

- anonymous

@mathstudent55 yes :) thank you :) I was just drawing it again and again but didn't think of the law of sines :)

- mathstudent55

I did not use the law of sines.
This is simply the definition of the sine in a right triangle.

- mathstudent55

|dw:1439439303636:dw|

- mathstudent55

We used the sine ratio which is opp/hyp.
|dw:1439439409054:dw|

- mathstudent55

To find the length of the beam, we did this:
|dw:1439439471920:dw|

- mathstudent55

To find the heigth the beam reaches at 60 deg, we did this:
|dw:1439439571183:dw|

- anonymous

wait, so like the beam will hit the wires at 60 something degree, because Sin^-1(53.89 over62.23 =59.99

- mathstudent55

It equals 63.91 deg.
Yes, when the beam is at 63.91 deg with the ground, the beam will hit the wires.

- anonymous

okay, thanks. i bad at round numbers

- mathstudent55

No problem.
You're welcome.

- Nnesha

cool ,-,

- radar

You beat me to this, @mathstudent55 and did an excellent step by step job. I could not done any better or clear explanation. You certainly deserve the medal.

- mathstudent55

@radar A good word from you means a lot to me. I really appreciate it. Also, thanks for the testimonial.

- anonymous

@mathstudent55 hey, so I got a mistake that the when th beam make angle of 40 degree then from the beam to the ground is 8ft, so the beam won't hit the wires since we know that the beam is 12,45 ( I had calculator again myself) and when it make angle of 60 then from the beam to the sruface is 10,78. by that, the beam WILL NOT hit the wires, so do I have to find anything else when I know that the beam will not hit the wires?

- radar

Please post the problem again, providing correct info. The problem as presently posted, states that at 40 degrees elevation, the top of the beam is 40 ft. above the ground. How you got 12.45 ft. is not clear.

- anonymous

okay

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