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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

Wow this looks complex

Is this a complete question?

Country A and country B exchange rate is 1:6, a product C is cost 11 dollars of country B currency.

Country A economy is in inflation, more than 10% of the money were issue, meanwhile social productivity increased by 20%, How many product C can be sold out by using 20 dollars of country A currency?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,232

hi NakulG

It does look complex. I didn't find the diagram helpful and some of the translation had me mystified for a while, so I made my own diagram:

I've shown lines at the front in thick ink, ones that would be hidden in thin ink and construction lines dotted. Alongside, I've drawn the cross-sectional shape. It is a kite.

Is the problem complete? Well you have enough information to make a wire model so I think the answer is YES.

Can I do it? Well let's see.

(i) AC is the projection of A'C onto the base, so that perpendicular property must follow as AC is perpendicular to BD.

(ii) To get the true angle between two intersecting planes first find the line of intersection. Then a line perpendicular to this in each plane. Then the angle between those lines.

In this case BD is the line of intersection. EC' and EA' are lines in the planes so find the angle C'EA'. You will need trigonometry to find the lengths of C'E, A'E and A'C', then use the cosine rule.

(iii) When the lines are in different planes you need to translate one line until it has crosses the other line, and then calculate the angle between those lines.

In this case translate AD across the base until the new A coincides with B. So draw a line BF, parallel to AD with F on DC. The angle required is angle FBC'. Again use trigonometry to find the lengths of the three sides in triangle FBC' and again the cosine rule.

Hope that helps.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

Thank you, this simplifies the problem completely... though i still need to calculate the answer...Thanks

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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

A'C = sqrt 19

A'D = sqrt 7

A'B = sqrt7

AE = 1

EC = 3

BE = DE = sqrt 3

Cos A'CA = 4 / (sqrt19)

Cos A'DA = 2 / (sqrt7)

Cos A'BA = 2 / (sqrt7)

On a separate note - in the question it is mentioned as a square prism, how is this a square prism? Any thoughts

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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

C'A = 2 sqrt7

C'E = Sqrt 21

CC' = 2 X Sqrt3

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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

A'C' = sqrt7

CF = Sqrt(3)

CC' = 2 sqrt(3)

FC' = Sqrt(15)

BC' = 2 sqrt(6)

triangle BCD = equilateral traingle with each side 2 sqrt(3)

a highly contorted figure...

*Last edited by NakulG (2015-03-07 11:11:29)*

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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

Angle C'EA' = arccosine (4.5/sqrt21) = 10.9 degrees (approximately)

*Last edited by NakulG (2015-03-07 17:08:47)*

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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

Angle FBC' = Arccos ((sqrt6)/4) = 52.2 degrees (approximately)

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**NakulG****Member**- Registered: 2014-09-02
- Posts: 186

thanks Bob your approach really helped ... a good question.

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