Harder 2 Unit Topic: Trigonometry (1 Viewer)

[XcarvengerX]

Just a couple of confusing questions:
1. Two interconnecting equilateral triangle-shape (note: not pyramids) buildings, ABD and BCE, are built on a horizontal plane so that AC is horizontal. AB is 4 metres and CE is 5 metres. Find the angle DE makes with the horizontal.
(Note: I managed to draw the diagram, but couldn't get the answers)

2. A circular hoop of radius 60 cm is suspended from a point by eight equal cords, each 1 metre in length. If the cords are equally spaced around the hoop, find the angle between two consecutive strings.
(Note: More like english essay than math to me)

Thank you for your help. Cheers

 

[airie]

I can't even understand the first question

For the second question, see diagram attached. I've only drawn the parts concerned, not the whole lot. And I'm being lazy here, just giving the method

Say A and B are two points of contact on the loop with two of the strings, and the loop is suspended by strings from point T. Connect centre of loop, O, with A and B by two imaginative lines, then since the strings are placed equally around the loop, angle AOB is 45 degrees. Then use cosine rule on triangle OAB to get the length of AB, and use cosine rule again, this time on triangle TAB, as you know the lengths of strings TA and TB are 1m, as well as the length of AB, to get angle ATB.

Hope that helps

 

[SoulSearcher]

Maybe if I saw the diagram for the first one, I could do it, because I'm currently fucked trying to draw it

 

[Riviet]

I don't get the "equilateral triangle shaped buildings interconnected" bit. The rest of the question just falls apart.

 

[XcarvengerX]

Say A and B are two points of contact on the loop with two of the strings, and the loop is suspended by strings from point T. Connect centre of loop, O, with A and B by two imaginative lines, then since the strings are placed equally around the loop, angle AOB is 45 degrees. Then use cosine rule on triangle OAB to get the length of AB, and use cosine rule again, this time on triangle TAB, as you know the lengths of strings TA and TB are 1m, as well as the length of AB, to get angle ATB.

Hope that helps

Got it, thanks. If you can tell me, what program did you use to make such a nice diagram? Paint?

For question 1, I am not sure, but I am going to scan my diagram when I can get to a scanner. Here is how I draw it. AC is a straight line with B somewhere in the middle. AB is 4 m and BC is 5 m. D is a point that makes ABD an equilateral triangle with 4 m for each side. E is a point that makes BCE an equilateral triangle with 5 m for each side. DE is connected and extended to AC (the horizontal). I don't know if this is right or not because I can't get to the answer.

 

[SoulSearcher]

Well I've got 10'54' for the first question, is that the answer?

 

[airie]

Got it, thanks. If you can tell me, what program did you use to make such a nice diagram? Paint?

Heh, yeah, Paint is the most convenient tool that also boots up the fastest

For question 1, I am not sure, but I am going to scan my diagram when I can get to a scanner. Here is how I draw it. AC is a straight line with B somewhere in the middle. AB is 4 m and BC is 5 m. D is a point that makes ABD an equilateral triangle with 4 m for each side. E is a point that makes BCE an equilateral triangle with 5 m for each side. DE is connected and extended to AC (the horizontal). I don't know if this is right or not because I can't get to the answer.

Well I've got 10'54' for the first question, is that the answer?

Got the same as SoulSearcher. Using cosine rule on triangle BDE to get DE=sqrt21, then sine rule to get angle DEB to be arcsin(2/sqrt7). As angles BEC and BCE are 60 degrees (equilateral triangles), 180 degrees minus the angle sum of DEC and ECB gives the angle DE makes with the horizontal. Is that right?

 

[SoulSearcher]

Got the same as SoulSearcher. Using cosine rule on triangle BDE to get DE=sqrt21, then sine rule to get angle DEB to be arcsin(2/sqrt7). As angles BEC and BCE are 60 degrees (equilateral triangles), 180 degrees minus the angle sum of DEC and ECB gives the angle DE makes with the horizontal. Is that right?

Yeah, that's pretty much my method, but don't know the answer, so can't say if we're correct right now.

 

[fishy89sg]

this isnt 2 unit, this is 3d trig 3 unit

 

[airie]

The first question doesn't look like 3D, judging from the diagram drawn

 

[SoulSearcher]

That's because it isn't 3d trig, hence part of harder 2 unit trig

 

[followme]

Yeah, that's pretty much my method, but don't know the answer, so can't say if we're correct right now.

Mmmm..I got your answer as well. Here is my working..
what did u guys get for question 2? close to 26'36'?

 

[Shivorken]

Question 1 was easy once i figured out how to draw the damn thing =]

10'54' just like soulsearcher got. I believe it should be correct.

Answers?

 

[Riviet]

Well if several people are getting the same answer, then it's highly likely to be correct.

 

[XcarvengerX]

Answer:
1. 17'22'
2. 26'33'

If the diagram is like what followme posted (thanks followme), then yeah I also got 10'54'. Either the answer or the diagram is wrong. Probably it is actually 3D somwhow.