VCE General & Further Maths Question Thread!

[keltingmeith]

Question relating to trigonometry and geometry.

The following diagram shows a cross-sectional view of a beehive with its regular hexagonal compartments. 
Only shows one triangle within a hexagonal compartment and the base at 10mm
Determine the value of an interior angle of each hexagonal compartment,  as shown in the diagram.

I put down 360/6 = 60 degrees,  but this was marked wrong.  How do you calculate this the right way?

I'd have to see the diagram to be absolutely sure, because if you have a regular hexagon, that's right.

[darklight]

Hi guys,

I worked out the right answer by using the sin rule, where the unknown (hypotenuse) is root 2 x over sin 60, the 'opposite' is x giving you E (if you let x initially be RT). However, considering that this could be 2 right-angled triangles, why can't you let x denote RT. Then from this you get RQ being root 2 x. As SR would also be x, then couldn't you use pythag to get SQ. However, this doesn't give you the same answer as the above (the correct one). Why not?

Hope this made sense! 

[plato]

Can someone help with this problem please? Ones in black pen thanks!

These should both be done with a TVM app on your calculator.
On the  TiNspire, it is called a Finance Solver (Menu-8)
On the Classpad, it is called Compound interest (Menu-Financial)

[plato]

Hi guys,

I worked out the right answer by using the sin rule, where the unknown (hypotenuse) is root 2 x over sin 60, the 'opposite' is x giving you E (if you let x initially be RT). However, considering that this could be 2 right-angled triangles, why can't you let x denote RT. Then from this you get RQ being root 2 x. As SR would also be x, then couldn't you use pythag to get SQ. However, this doesn't give you the same answer as the above (the correct one). Why not?

Hope this made sense! 

First up, don't worry about using x. Just let RT = 1 unit long.
The problem is that SR does not equal 1 (or x in your case)

Because angle RQT = 45 degrees, QT = 1 unit
By Pythagoras,

In triangle SRQ, angle SQR = 30 degrees because angle PQR = 90 and angle PQS = 60.

Then
So 

[darklight]

Hey guys,

Not sure how to answer this, e.g. for 12 a)i I get 360 m, answer is 362m.
Isn't it just 2/2.4 = 300/x, or is this an oversimplification of the problem?

[~V]

Hey guys,

Not sure how to answer this, e.g. for 12 a)i I get 360 m, answer is 362m.
Isn't it just 2/2.4 = 300/x, or is this an oversimplification of the problem?

I asked my teacher about this one, we did it a couple of times and in the end she told me it was an error from the book... but i did get 360m over and over again.

[plato]

Hey guys,

Not sure how to answer this, e.g. for 12 a)i I get 360 m, answer is 362m.
Isn't it just 2/2.4 = 300/x, or is this an oversimplification of the problem?

The view 300m from the window is 302m from where the girl is standing. If you use this distance, the answer is 362.

[darklight]

Hi guys,

If this is all the information that is given, would you be able to work out ST or not? Or do they have to give you angle SQT? Could you for example make a triangle PST and work it out from there?

[InsaneMDot]

Hi guys,

If this is all the information that is given, would you be able to work out ST or not? Or do they have to give you angle SQT? Could you for example make a triangle PST and work it out from there?

I think since angles SQP and TQP are 90 degrees, SQT has to be 90 degrees?
So a pythagoras calculator should work.

[keltingmeith]

I think since angles SQP and TQP are 90 degrees, SQT has to be 90 degrees?
So a pythagoras calculator should work.

Nope - we're working three dimensions here, and so the angle SQT could be anything.

I've been racking my brain, and I can't find a plausible way to find the length ST using knowledge from high school geometry. You sure there's nothing else there?

[Zlatan]

Nope - we're working three dimensions here, and so the angle SQT could be anything.

Could you explain how you know that pythag would not work in this situation ??

Thanks

[keltingmeith]

Pythag only works for right handed triangles.

We don't have any of the angles for either the triangle SQT or the triangle SPT, so we cannot confirm that they're right-angled triangles, and so we cannot use Pythag. Using Pythagoras *could* work, but we don't have enough information about the situation to know if it will work.

[darklight]

It was just something I was wondering about
But could this problem we tackled with higher maths skills? Intrigued what the answer would be.

[keltingmeith]

I haven't tried it at all, but I suspect that you could use linear algebra.

If you turn each line-segment into a vector, then this layout as we see would exist in . From this, we can see that QP is the result of QS cross QT. If you define Q as the origin of our vector space, you can then set QS as (a, b, c), QT as (d, e, f) and QP as (0, 0, g), you could then use the varying vectors/line segments to find a, b, c, d, e, f and g, and then you would be able to find the angle SQT, which would mean you can use the cosine rule to find ST or simply find the length of the vector ST from basic vector addition.

Might not have enough information though - this is all conjecture, I haven't actually tried to evaluate it, nor do I have the time to try.

[engton1796]

I'd have to see the diagram to be absolutely sure, because if you have a regular hexagon, that's right.

Here.  My teacher,  as you can see,  didn't give me a mark.  The use of this 60 degree angle in other questions made my other questions correct but not this?