| Author |
Message |
Lan Pham
science forum beginner
Joined: 18 Jun 2005
Posts: 3
|
 Posted: Sat Jun 18, 2005 3:32 am Post subject:
please help
|
|
|
Hi all,
i need help, here the question:
1. From a point on the ground the angle of elevation to a ledge on a building is 27°, and the distance to the base of the building is 45 meters. How many meters high is the ledge?
meters high is the ledge?
the correct answer is 45/tg 27 degree. I dont know why? pleaee help. thanks |
|
| Back to top |
|
 |
Lost Maths Worker
science forum beginner
Joined: 03 May 2005
Posts: 18
|
| Posted: Sat Jun 18, 2005 4:38 am Post subject:
Re: please help
|
|
|
45cos27 = x
where x = height
x = around 40.10m |
|
| Back to top |
|
|
Jeremy Watts
science forum Guru Wannabe
Joined: 24 Mar 2005
Posts: 239
|
| Posted: Sat Jun 18, 2005 7:50 am Post subject:
Re: please help
|
|
|
"Lan Pham" <Lanpham26@yahoo.com> wrote in message
news:1719877.1119065584627.JavaMail.jakarta@nitrogen.mathforum.org...
| Quote: | Hi all,
i need help, here the question:
1. From a point on the ground the angle of elevation to a ledge on a
building is 27°, and the distance to the base of the building is 45
meters. How many meters high is the ledge?
meters high is the ledge?
|
what do you know about trigonometry? the tan of an angle is defined as the
'opposite' / 'adjacent' . Here the 'opposite' will be the height of the
ledge, the 'adjacent' the distance from the observer to the building, you
can do the rest
| Quote: |
the correct answer is 45/tg 27 degree. I dont know why? pleaee help.
thanks |
|
|
| Back to top |
|
|
Anon
science forum Guru Wannabe
Joined: 03 Jun 2005
Posts: 184
|
| Posted: Sat Jun 18, 2005 12:56 pm Post subject:
Re: please help
|
|
|
On Fri, 17 Jun 2005 23:32:34 EDT, Lan Pham <Lanpham26@yahoo.com>
wrote:
| Quote: | Hi all,
i need help, here the question:
1. From a point on the ground the angle of elevation to a ledge on
a building is 27°, and the distance to the base of the building is
45 meters. How many meters high is the ledge?
the correct answer is 45/tg 27 degree. I dont know why? pleaee help. thanks
|
Not correct.
tangent( 27 degrees ) = 'opposite' (height of ledge) divided by
'adjacent' (distance to base of building)
So tan(27 degrees) = h / 45, or
h = 45 x tan(27 degrees), in meters |
|
| Back to top |
|
|
Guest
|
| Posted: Sat Jun 18, 2005 2:32 pm Post subject:
Re: please help
|
|
|
Lost Maths Worker wrote:
| Quote: | 45cos27 = x
where x = height
x = around 40.10m
|
No, x = 45*tan(27). |
|
| Back to top |
|
|
Barry Schwarz
science forum beginner
Joined: 30 Apr 2005
Posts: 31
|
| Posted: Sat Jun 18, 2005 4:05 pm Post subject:
Re: please help
|
|
|
On Fri, 17 Jun 2005 23:32:34 EDT, Lan Pham <Lanpham26@yahoo.com>
wrote:
| Quote: | Hi all,
i need help, here the question:
1. From a point on the ground the angle of elevation to a ledge on a building is 27°, and the distance to the base of the building is 45 meters. How many meters high is the ledge?
meters high is the ledge?
the correct answer is 45/tg 27 degree. I dont know why? pleaee help. thanks
If the building is vertical, then you have right triangle. The |
definition of tangent is opposite side over adjacent side. The
opposite side is the height of the ledge, call it h. The adjacent
side is 35 meters. Therefore tan(27) = h/45 and h = 45*tan(27).
45/tan(27) is incorrect.
<<Remove the del for email>> |
|
| Back to top |
|
|
Lan Pham
science forum beginner
Joined: 18 Jun 2005
Posts: 3
|
| Posted: Sun Jun 19, 2005 2:52 am Post subject:
Re: please help
|
|
|
yes
im sorry, the correct answer is 45*tg(27)
thank all |
|
| Back to top |
|
|
Google
|
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
Forum index
»
Science and Technology
»
Math
»
Undergraduate
