can any1 help with 3-Dimensional trig????????? (1 Viewer)

[pink_rabbit_<3]

i find trig the hardest of the extension 1 topics!! OK these 3 questions are killin me..i cant just figure out how to get the answer.. can any1 help plz??

1. David walks along a straight road. At one point he notices a tower on a bearing of 053 degrees with an angle of elevation of 21 degrees. After walking 230 m, the tower is on a bearing of 342 degrees, with an anglel of elevation of 26 degrees. Find the height of teh tower correct to the nearest metre.

Answer: 84m

2. A cylinder with a radius of 4cm and perpendicular height 15cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?

Answer: 16 degrees 20 '

3. A hot air balloon flying at 950m/h at a constant altitude of 3000m is observed to have an angle of elevation of 78 degrees. After 20 minutes, the angle of elevation is 73 degrees. Calculate the angle through which the observer has turned during those 20 mintues.

Answer: 11 degrees 10''

 

[SoulSearcher]

The answer to question 2 is in this thread, although for some reason the answer is 16'50' in that thread:
http://community.boredofstudies.org/13/mathematics-extension-1/108606/3d-trig-question.html

 

[SoulSearcher]

1) Let h be the height of the tower
Let x and y be the length of the sides of the triangle lying flat on the ground
h/x = tan 21
.'. x = h cot 21
h/y = tan 26
.'. y = h cot 26
Using the cosine rule,
230² = h²cot²21 + h²cot²⁶ - 2h²cot 21 * cot 26 * cos 71
.'. 52900 = h²(cot²21 + cot²26 - 2cot 21 * cot 26 * cos 71)
.'. h² = 52900/(cot²21 + cot²26 - 2cot 21 * cot 26 * cos 71)
= 7041.719265...
.'. h = 83.91495257...
= 84 metres correct to the nearest metre

 

[AFGHAN22]

yo its me, ask me tomorrow and ill give it a go

 

[Wackedupwacko]

3) ummm without a drawn diagram its kinda hard to explain to u umm but ill give ur 3 triangles u should have. first one is right hand incline 78degrees with a height of 3000 and hypotenuse x. 2nd one is right hand with and incline of 73degrees and 3000 height and hypotenuse y. ur 3rd triangle should be just a scalene with one side being 950/3 (distance travelled by balloon) one side x and one side y. the angle between the x and y lines is the one u want let it be @

with that its pretty simple
x = 3000/sin78
y = 3000/sin73

now applying the cos rule in the 3rd triangle and ull get @. which is how much the observer turned.. its kinda late so im not going to do the exact numbers.

explanation of the triangles the first one is starting triangle. 2nd one is ending. third one is basically a top view of it. if u want a 3d diagram u shold draw urself a rectangle first with lengths 950/3 and height 3000 then put a point somewhere just below the rectangle and connect the point to all cornors of the rectangle label one angle of elevation 78 degrees and the other one 73degrees.

 

[pink_rabbit_<3]

thanks guys !!!! your help is greatly appreciated!! =)
hmmm yeh i think the answer in my text book is wrong for question 2...

 

[~shinigami~]

1) Let h be the height of the tower
Let x and y be the length of the sides of the triangle lying flat on the ground
h/x = tan 21
.'. x = h cot 21
h/y = tan 26
.'. y = h cot 26
Using the cosine rule,
230² = h²cot²21 + h²cot²⁶ - 2h²cot 21 * cot 26 * cos 71
.'. 52900 = h²(cot²21 + cot²26 - 2cot 21 * cot 26 * cos 71)
.'. h² = 52900/(cot²21 + cot²26 - 2cot 21 * cot 26 * cos 71)
= 7041.719265...
.'. h = 83.91495257...
= 84 metres correct to the nearest metre

Wow SoulSeacher, you beast!

My grasp of 3-D trig sucks hard compared to you.

 

[SoulSearcher]

Wow SoulSeacher, you beast!

My grasp of 3-D trig sucks hard compared to you.

I'm pretty sure that you should be able to do these kinds of questions next year with no trouble at all

 

[GaDaMIt]

I'm pretty sure that you should be able to do these kinds of questions next year with no trouble at all

lol