building heigh differences via trigonometry  TOPIC_SOLVED

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building heigh differences via trigonometry

Postby etxei on Fri Aug 19, 2011 1:02 pm

i have attempted the following problem but my answers differs from authors and would apreciate if any1 can assist, i do not need the exact method but only to know wether my method is utterly wrong...

standing at the top of a 150ft bulding , an observer angle of depression to the top of a smaller building is 12 degrees, and 34 degrees to the base of the smaller building, thus find the height of the smaller building.

my answer give the smaller building heigh as 120 ft, whereas by the authors if 102.7 ft

solved it by tan 46=opposite/adjacent= tan 46=150/adjacent= solving for the adjacent =144.85

then using the adjacent with the triangle with the vertex at 12 degrees i attempted to fin the opposite which i could use to deduct from the 150 ft to obtain the difference in height.

this give tan 12=opposite/144.85 solving for the opposite=30.78 ft

which i deduct from the height of the tallest building to obtain the difference of height...but so far no joy
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Postby stapel_eliz on Fri Aug 19, 2011 4:58 pm

etxei wrote:standing at the top of a 150ft bulding , an observer angle of depression to the top of a smaller building is 12 degrees, and 34 degrees to the base of the smaller building, thus find the height of the smaller building.

my answer give the smaller building heigh as 120 ft, whereas by the authors if 102.7 ft

solved it by tan 46=opposite/adjacent= tan 46=150/adjacent= solving for the adjacent =144.85

Where is "46" coming from? What, exactly, does "adjacent" stand for, in terms of the two buildings and / or the distance between them? :wink:
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Re: building heigh differences via trigonometry

Postby etxei on Fri Aug 19, 2011 7:35 pm

thanks for the interest to start with...

effectively adjacent stand for the distance between the buildings, 46 degrees comes from adding both 12 and 34....

if you were to draw a vertical line, say 200 ft long, 150 of those representing the total lenght of the tallest building, then a x distance away draw a another vertical line, smaller of course than the first line. the first angle of depresion standing on the tall building roof will be 12 degrees, ( down to the roof of the smaller building), the second angle of depresion ( to the base of the smaller) will be 34. i have assumed that maybe life will be easier if i treated it as a right triangle with the opposite lenght being 150 ft, and the angle as 12+34=46. and the adjacent been the distance between the buildings

hope it makes a bit more sense now..cheers
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Postby stapel_eliz on Fri Aug 19, 2011 8:21 pm

etxei wrote:effectively adjacent stand for the distance between the buildings, 46 degrees comes from adding both 12 and 34....

You might want to draw a picture. You have two angles of depression, with the smaller "contained within" the larger. (You'll see then when you draw the picture with the two lines of sight from the top of the taller building.) :wink:
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Re: building heigh differences via trigonometry

Postby etxei on Fri Aug 19, 2011 9:52 pm

http://www.mecmath.net/trig/trigbook.pdf

hope you have the adobe reader, is the first exercise on page 20....
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Postby stapel_eliz on Sat Aug 20, 2011 12:01 am

etxei wrote:http://www.mecmath.net/trig/trigbook.pdf

hope you have the adobe reader, is the first exercise on page 20....

What were you wanting us to see? The picture they have is exactly what you should have drawn and, as the picture shows, the twelve-degree angle is "inside" the thirty-four-degree angle, so adding made no sense.

Instead, try setting things up using the picture they've given you, with, say, a distance of "x" between the two buildings, and the additional height (for the twelve-degree-angle triangle "above" the smaller building) of 150 - h, where "h" is the height you're trying to find. :wink:
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Re: building heigh differences via trigonometry  TOPIC_SOLVED

Postby little_dragon on Sat Aug 20, 2011 1:21 am

do like they said:
tan(12)=(150-x)/h
tan(34)=x/h
solve for h
:wave:
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