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Mimi_x3
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An electricity tower is h metres high. The elevation angles of the tower from Aaron and Kevin are respectively 45° and 30° . On the ground, relative to the tower the bearings of Aaron and Kevin are respectively SW and SE. If the boys are 50m apart, find the height h of the tower to the nearest ontenth of metre.
Can someone please explain, I'm so lost xD
 3 years ago
 3 years ago
Mimi_x3 Group Title
An electricity tower is h metres high. The elevation angles of the tower from Aaron and Kevin are respectively 45° and 30° . On the ground, relative to the tower the bearings of Aaron and Kevin are respectively SW and SE. If the boys are 50m apart, find the height h of the tower to the nearest ontenth of metre. Can someone please explain, I'm so lost xD
 3 years ago
 3 years ago

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Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
Do i work it out by soh cah toa ?
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
Okay, btw I asked the question before and someone drew a different diagram to yours xD
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
this problem is 3 dimensionaldw:1316171303379:dw
 3 years ago

raheen Group TitleBest ResponseYou've already chosen the best response.0
you are right.
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
from first two diagrams we get AT = h and KT = \[h \sqrt{3}\] so from 3rd figure AT2 + KT2 = 2500 so on simplifying h = 25 m
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
the first 2 diagrams are 2 dimensional and the 3rd figure is an overhead view
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
ohh I get it tyvm (:
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
dw:1316171761134:dw Is this right as well ?>
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
but the two lines below are inclined at 90 degrees to each other in the 3rd dimension
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
So its wrong ?
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
the figure is somewhat like .... try cutting a pyramid with a square base from one corner of base to it's opposite corner symmetrically. then u get a figure similar to what we need (not exactly similar but good enough to understand the above problem)
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
So the figure I drew was wrong ?:
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
dw:1316172253129:dw
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
oh 3D trig ?
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
the above figure is possible only in 3d
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
it can be in 3D ?
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
no no u don't need 3d trig. just consider each triangle at one time to get the required variables
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
Okay, but is it necessary though to draw it ?
 3 years ago

abhinayreddy Group TitleBest ResponseYou've already chosen the best response.2
the whole point is to test whether u can imagine the required diagram once u get the diagram it is easy simplification. u can see that i used the first two diagrams to get the values of AT and KT in terms of h and then used pythagoras theorem in the final triangle. drawing the diagram is important unless u can visualize it exactly to solve the problem in ur head
 3 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
Okay tyvm (:
 3 years ago
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