Solution of triangle

[Vastor]

Hey guys,

 

In the figure above, ABC and APQ are two triangles such that PQ is parallel to BC. Given that BP = 2cm, CQ = 4cm, BC = 6cm and PQ:BC = 1:3. Find

(a) angle of BAC

(b) the ratio of the area of [math] \Delta APQ [/math] to the area of trapezium BCQP.

 

for (a). Well, I'm assumed that AP:AB = 1:3 = AQ:QC beforehand, but the calculation doesn't work out. (Answer = 75"31')

 

and what I learn on this chapter is sine rule, cosine rule, and area of triangle based on these rule. so, can anyone give some hint about the trick behind this question because non of the method given can be used!

 

thnx.

Edited April 10, 2012 by Vastor

[Axioms]

Did you find the ratio between AP and PB? You can then get a total length for AB. Do the same on the other side and you can then apply the cos rule.

[Vastor]

Did you find the ratio between AP and PB? You can then get a total length for AB. Do the same on the other side and you can then apply the cos rule.

 

that's what I'm confuse about... I don't have any idea about (on how to obtain) the ratio of AP and PB so that I can use cos rule. Thus leave me with only length of 1 of the side(BC).

[imatfaal]

Was your answer 75.5 degrees or is that the correct one? Cos I get it too! I was just using rules on similar triangles (the 3 angles on ABC are same as APQ).

 

Show your working and we will see what is wrong.

[Xittenn]

that's what I'm confuse about... I don't have any idea about (on how to obtain) the ratio of AP and PB so that I can use cos rule. Thus leave me with only length of 1 of the side(BC).

 

 

You have your unknown AP call it x so what is AB equal to? You have side c for both triangles so it only makes sense to use similar triangles, and ratio and proportion. Cos this all makes so much sense 75.5 quickly becomes a very appropriate answer . . . .

[J.Nassar]

Try using the Theorem of Al-Kashi, you would get this: BC²=BA²+CA² - (2.BA.CA.cosBAC)

 

Sorry, my answer is very late but since no one answered...

Edited April 14, 2012 by Formicidae

[Vastro]

hey guys,

 

thank you for reply, I'm just having some bad moment with my account.

 

anyway, I think I just do some little mistake, btw, Imatfaal, that's the real answer.

[Vastro]

Hey guys,

 

 

 

15. In the figure above, AB is parallel to DC. AC and BD intersect at the point E. Given that AD = 6 cm, AC = 8cm and the (area of BDC = 12 cm^2). Find angle of DAC.

 

well, I don't know which 'trick' I can use to solve this question. I mean I found that:

 

[math] \Delta BDC = 12 = \frac{1}{2} * BD * DC * sin BDC = \frac{1}{2} * DC * BC * sin DCB = \frac{1}{2} * BD * BC * sin CBD [/math]

 

but then, I don't know where to start!

[imatfaal]

Vastor,

 

OK you need to use two ways of working out the area of a triangle.

 

1. Half base times height

2. Half length of two sides times the sine of the interior angle.

 

Which if you think of it are the same - just a different perspective.

 

Hint 1 AB and DC are parallel - - look to method one above

 

Hint 2 What do you know about the triangle DAC ? What other piece of info will allow you to get the needed info -- look to method 2 above

 

Try using the Theorem of Al-Kashi, you would get this: BC²=BA²+CA² - (2.BA.CA.cosBAC)

 

Sorry, my answer is very late but since no one answered...

 

Formicidae - we try not to give explicit answers here. This is a homework forum and the idea is to give hints, guide in the right direction, and correct mistakes - rather than give an answer. I am sure you appreciate that one just doesn't learn without doing the work

 

And I learnt from your post about al-Kashi, who I didn't know about before - so Thanks

 

http://en.wikipedia.org/wiki/Jamsh%C4%ABd_al-K%C4%81sh%C4%AB

[Vastor]

Vastor,

 

OK you need to use two ways of working out the area of a triangle.

 

1. Half base times height

2. Half length of two sides times the sine of the interior angle.

 

Which if you think of it are the same - just a different perspective.

 

correct me if I'm wrong but doesn't the (2) method is just derived from (1) method

 

[math] \Delta ABC = \frac{1}{2} * BC * AB * sin B [/math]

 

[math] \Delta ABC = \frac{1}{2} * BC * AB * \frac{h}{AB} [/math] where h = height

 

[math] \Delta ABC = \frac{1}{2} * BC * h [/math]

 

 

 

Hint 1 AB and DC are parallel - - look to method one above

 

I don't get this (as a) hint. I mean the "h" is unknown too. there too much unknown already (look below picture)

 

 

 

Hint 2 What do you know about the triangle DAC ? What other piece of info will allow you to get the needed info -- look to method 2 above

 

what I know about triangle DAC? not much, maybe you can know it by look at my messy perspective.

 

 

 

well let's see:

1. The angle of A = angle of B = k1

2. (assumption) that AD = BC and AC = BD (and always?)

 

uh, forgot to say, I got the answer already (30 degree). From my assumption, but I doesn't know it will always happen like that because I can't find logical reasoning behind it. I mean you can move B further, and the only thing that change is angle, thus making it length equal no more!

 

btw, for (1), from this point, if BD and BC are known, we can get answer by using your (2) method.

[Joatmon]

Help! Last time I looked at this post I am sure that I saw the diagram. Now I only see what I attach . Would anyone like to tell me the kind of picture it is and what it should be opened with? I am not aware of making any changes to my computer.

[Joatmon]

Help! Last time I looked at this post I am sure that I saw the diagram. Now I only see what I attach . Would anyone like to tell me the kind of picture it is and what it should be opened with? I am not aware of making any changes to my computer.

 

Later - just had another look and I now have the diagram - anyone know why I lost it earlier?

 

 

 

[imatfaal]

Vastor - you are overworking and underthinking the problem.

 

You have the area of triangle DBC - now look at triangle DAC; whilst doing this keep in mind half base times height. Now what do you know about triangle DAC?

 

Using the new info about triangle DAC - now think about the area of a triangle being half length two sides times the sine of the interior angle.

[Vastor]

Vastor - you are overworking and underthinking the problem.

 

You have the area of triangle DBC - now look at triangle DAC; whilst doing this keep in mind half base times height. Now what do you know about triangle DAC?

 

Using the new info about triangle DAC - now think about the area of a triangle being half length two sides times the sine of the interior angle.

 

what I know about DAC? diagram below and the stuff that I post before!

 

 

try my best(thinking for hours), and the best I got is

 

[math] 12 = \frac{1}{2}*(AC)*(H) = \frac{1}{2}*(BD)*(BC)*(sin DBC) [/math]

 

I don't know what kind of 'logical approach' you are trying to use, but the sure thing is I can't grasp it at all. D:

Edited April 18, 2012 by Vastor

[imatfaal]

Vastor,

 

you are going to kick yourself.

 

Area of Triangle is half base times perp height

Triangle DBC - base is DC and perp height is distance between lines AB and DC

Triangle DAC - base is DC and perp height is distance between lines AB and DC

Thus area of triangle DAC is the SAME as triangle DBC - ie 12 cm^2

 

area of triangle is also product of two sides and the sine of the interior angle ie

[math] Area = \frac{1}{2}ab\sin C [/math]

or with our figures for triangle DAC

[math] Area =12 = \frac{1}{2}AD.AC\sin DAC [/math]

 

You know AD, AC and the Area - you thus have only one unknown, rearrange and you are done

[Vastor]

Vastor,

 

you are going to kick yourself.

 

Area of Triangle is half base times perp height

Triangle DBC - base is DC and perp height is distance between lines AB and DC

Triangle DAC - base is DC and perp height is distance between lines AB and DC

Thus area of triangle DAC is the SAME as triangle DBC - ie 12 cm^2

 

area of triangle is also product of two sides and the sine of the interior angle ie

[math] Area = \frac{1}{2}ab\sin C [/math]

or with our figures for triangle DAC

[math] Area =12 = \frac{1}{2}AD.AC\sin DAC [/math]

 

You know AD, AC and the Area - you thus have only one unknown, rearrange and you are done

 

wow, never see that coming!

 

so, basically, this would answer my wonder about this shape also that:

 

- area of triangle AED = area of triangle BEC; because triangle EDC is the "intersection" between the triangle (and obviously both has same amount of area)

 

- 'every' kind of this shape have same amount of area, right?; I mean:

at first, when I thought if we move the "B" only, the shape will distort thus make both triangle have different area, etc..

but then, "E"(the intersection line) will changed also for keeping the shape. Thus make both triangle have equal area whatsoever!

 

thank you very much.

 

P.S. my ambition is to become a theoretical physicist/mathematician. Overlooking the answer of such simple "puzzle" give me a bad feeling! D; *kicking myself*

[imatfaal]

wow, never see that coming!

Once you see it - all becomes clear.

 

so, basically, this would answer my wonder about this shape also that:

 

- area of triangle AED = area of triangle BEC; because triangle EDC is the "intersection" between the triangle (and obviously both has same amount of area)

Yes - I had to check that for a second - but you are correct

 

- 'every' kind of this shape have same amount of area, right?; I mean:

at first, when I thought if we move the "B" only, the shape will distort thus make both triangle have different area, etc..

but then, "E"(the intersection line) will changed also for keeping the shape. Thus make both triangle have equal area whatsoever!

It depends on lines AB and DC being parallel and shared base - any set of points on parallel lines should work out the same (try a few and make sure)

 

P.S. my ambition is to become a theoretical physicist/mathematician. Overlooking the answer of such simple "puzzle" give me a bad feeling! D; *kicking myself*

Don't be silly - most of it is practice and repetition; which can only come with age. Don't worry about it - learn from it. Your mistake was to not delve into what the information the question gave you really meant - and once you had got past the initial thoughts on those lines it is very hard to start afresh.