ABC is a triangle such that angleB=90 degre, /AB/=3x, /BC/=y. Where 'y' is less than 4x. BC is produced to D.such that /BD/=4x. a. If N is the foot of the perpendicular from C to /AD/ prove that triangle CND is similar to triangle ABD. b. Find in terms of x and y expresion for /CN/ and /AN/

OMG. Where are the mathematicians in d house

Hi, I saw your question about 2-3am in the morning. I got interested in it, and so I decided to give it a try. I last did something like this more than 5 years ago! But I guess I have one or two ideas that can help you solve the problem. I hope my response is not late sha,

Simonwal:

ABC is a triangle such that angleB=90 degre, /AB/=3x, /BC/=y. Where 'y' is less than 4x. BC is produced to D.such that /BD/=4x. a. If N is the foot of the perpendicular from C to /AD/ prove that triangle CND is similar to triangle ABD. b. Find in terms of x and y expresion for /CN/ and /AN/

Now, we say two triangles are similar if they have the same corresponding angles. To solve the question

1. Draw a right angled triangle ABC. Put the right angle on B.
2. From point C, draw a horizontal line and stop at a point which you should call D.
3. Now draw a slant line from the point A to your new point D. You will notice that you have two triangles with two slants i.e. slant AC and slant AD.
4. From point C, draw a vertical line and stop when you get to a point on the slant AD. Call that point N; that is, N is the point where you stop the vertical line on slant AD. I hope you are following. It is a vertical line you should draw here, not anything that is bendy
5. Now put angle 90 (right angle) at C because C is in fact the point from where you started drawing the vertical line. You are almost done with the solution
6.It should be clear that you have some triangles all inside one another. But be careful.
7. Take out the small triangle CND. That is the first triangle, and it is very useful for your proof
8. Take out the big triangle ABD, which looks like a house that houses the other triangles.
9. So you now have triangle CND(small one) and triangle ABD (big one). Good!
10. Retain the right angle at B (i.e. angle 90) for the big triangle. Also, retain the right angle at C (i.e angle 90) for the small triangle.
11. After you have done that, you will notice that angle D appears on both triangles. Call the angle theta or D or whatever.

Proof proper

Now that you have two triangles: small one CND (where C is right angle) and big one ABD (where B is right angle), you can prove that triangle CND is similar to triangle ABD by just recognizing that: angle B=angle C=90 degrees(right angle), angle D in small triangle=angle D in big triangle and, lastly, angle A=angle N because angle A=90-angle D and angle N=90-angle D. They are the same. This proves that the triangles are similar and therefore their sides must be in constant ratio-

Second question

Just use the result from the proof. Since the triangles are similar, then their corresponding sides must be in a constant ratio. But before you use this fact, you need to first determine AD in the big triangle. How do you do this? Well, just use Pythagorean rule. Because BD=4x and AB=3x, and B=90 (right angle), then it follows that AD=5x (prove this). With this, you have completely characterized the big triangle. You now know all its sides--hahahah good.

Now, you need to tackle the small triangle. This is where you will need to use the similar triangle rule I stated above: If two triangles are similar, their corresponding sides must be in the same ratio. So,

AB/CN = BD/CD. This gives: 3x/CN=4x/(4x-y)---------you can then make CN the subject of the formula. Remember, CD=BD-BC=4x-y (look at the main triangle again)

AN is also easy. AN=AD-ND (look at the main triangle again). You know AD (as calculated above), but you need to find ND. So, what is ND?

Well, you can find ND from the small triangle! You know we have found CN ( see above). Use CN and CD to find ND by Pythagorean rule. This is possible because the small triangle is right angles at C (remember? go over the above explanation again if you don't remember). So, ND= SQRT(CN^2+CD^2).

Finally, AN=AD-ND=5x minus the answer you got for ND.

Hope these long explanations help?

Ask questions if you need further help.

T22

Good morning bro. You are indeed a guru. Many thanks to u. If i have any problem i wil let u know. Once again thank you

Pls help me out on this.:A loan requires d borrower to repay #1000 after 1year,2000 after 2yr,3000 after 3yr and 4000 after 4yr. At what time wil he pay a single payment of 10000 to b re-pay alone wit 4% effective rate

I never liked Maths and I don't think I'd ever like it, at least not after seeing this, maybe in my next life sha. See Equation, I don lost

sincerely speaking this guy is a guru. He solved the problem correctly. Thanx you very much.

That thing simple na,una for leave am for me to come and solve.

Occupy Nairaland Occupy Nairaland Occupy Nairaland
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Occupy Nairaland Occupy Nairaland Occupy Nairaland
Occupy Nairaland Occupy Nairaland Occupy Nairaland
Occupy Nairaland Occupy Nairaland Occupy Nairaland[/quote]

Folons: Pls help me out on this.:A loan requires d borrower to repay #1000 after 1year,2000 after 2yr,3000 after 3yr and 4000 after 4yr. At what time wil he pay a single payment of 10000 to b re-pay alone wit 4% effective rate

I think You need to use the formula for calculating Annuity A = P(1+R/100)n (the n is superscript)

It is very easy to be drawn into circular arguments when proving this algebraically as you only have one expression for one of the sides in triangle CND but two variables (x and y). To prevent this, you have to use two different ratios:

(1.) Ratio between similar sides from the two different triangles
(2.) Ratio between adjacent/opposite in the same triangle.

If your value for the length of the sides simultaneously satisfies (1.) and (2.), then your proof is validated.

See working out below:

Angles are always three in number in a triangle and they sum up to be 180.
Tri(ABD) and Tri(CDN) share the same Angle(D) = Ang(CDN) = Ang(BDA).
Also, both triangles[Tri(ABD) and Tri(CDN)] are right-angled triangle, this makes them to share simillar angles =90.
As a result this makes Ang(BAD) and Ang(NCD) complimentary angles to Angle(D) = Ang(CDN) = Ang(BDA); thus making them [Ang(BAD) and Ang(NCD)] equal to each other.
Therefore, Tri(ABD) and Tri(CDN) are simillar triangles.

/AB/ = 3x, /BC/ = y, /BD/ = 4x
By pythagoras rule, /AD/ = sqrt(/AB/² + /BD/²) = 5x
/CD/ = /BD/ - /BC/ = 4x-y

Considering Tri(CDN), by sine rule,
[sin <(CDN)]/[/CN/] = [sin <(CND)]/[/CD/] = [sin <(NCD)]/[/ND/]
Note: <(CND) = 90^o and <(CDN) = arcsin [(/AB/)/(/AD/)] = arcsin (3/5)

/CN/ = [3*(4x-y)]/5

/ND/ = [4*(4x-y)]/5
/AN/ = /AD/ - /ND/ = (9x+4y)/5

^^^ QED

Nice questions i will like to try all the questions, truly this can help Jambites/jambitos we can just share our problems here may be i will create a forum for that or what do you all think.

saintneo:

Considering Tri(CDN), by sine rule,
/CN/ = [3*(4x-y)]/5

/ND/ = [4*(4x-y)]/5
/AN/ = /AD/ - /ND/ = (9x+4y)/5

Apart from the fact that he should not be introducing a third variable (sin(unknown angle)), he should not be using sine rule here, neither is this the sine rule

Also:

CN/sin D ≠ AB/sin D

freecocoa: That thing simple na,una for leave am for me to come and solve.

same here!

freecocoa: That thing simple na,una for leave am for me to come and solve.

same here!

freecocoa: That thing simple na,una for leave am for me to come and solve.

same here!

Ibime:

Apart from the fact that he should not be introducing a third variable (sin(unknown angle)), he should not be using sine rule here, neither is this the sine rule

Also:

CN/sin D ≠ AB/sin D

Ibime, you should look deep. My apologies for not elaborating my methods.

by sine rule:
[sin <(CDN)]/[/CN/] = [sin <(CND)]/[/CD/] = [sin <(NCD)]/[/ND/]
Note: <(CND) = 90^o and <(CDN) = arcsin [(/AB/)/(/AD/)] = arcsin (3/5)

freecocoa: That thing simple na,una for leave am for me to come and solve.

You can as well solve the 2nd question on compound interest

freecocoa: That thing simple na,una for leave am for me to come and solve.

see ur head, na after them don solve am finish na im u wake up... U de try well well
meanwhile, what's with occupy nairaland'?

arapoa:
same here!

make una solve the second one instead of making noise!

Ibime and T22’s interpretations of the information are nearly correct, but Saintneo’s interpretation is completely in line with the information given. Proofs to both interpretations have been given, but full marks can only be awarded for correct interpretations and correct proof. It’s been nearly 20 years since I did basic maths, but here we go.

It noteworthy that the foot of a perpendicular line is the point that it makes a right angle (90’). Therefore, N is located on line A – D. (just like Saintneo’s drawing)

Proof:

Inverting CND so that it looks like ABD. (You don’t have to invert, it just makes it easier to the eyes. Doing this is quite elementary and I would simply go straight to prove the existing interpretation (drawing) of the question)

In the drawings call the angles at ABD a, b and d and do the same for CND i.e. c, n and d.

Angles n and b are both right angles = 90’
Therefore n = b.
Angle d is the same for both triangles.

Sum of angles in a triangle = 180’
Therefore Angle c = 180 – n – d = 180 – 90 – d = 90 – d.
Similarly Angle a = 180 – 90 – d = 90 –d.
Therefore Angle a = Angle c

2.

A-B = 3x, B-C = y, B-D= 4x
using Pythagoras rule.

A-D = Sq root of(A-B)sq + (B-D)sq = 5x
C-D = B-D – B-C = 4x-y

SOHCAHTOA:

In ABD sides = 5x, 3x and 4x.

In NCD sides = 4x-y, ND and NC.

Angle d is the same for both triangles: lets go,

In ABD sin d = 3x/5x = 3/5 ------- (1)

In NCD sin d = NC/4x – y ------ (2)

(1)= (2) thus:

3/5 = NC/4x-y
NC = 3(4x-y)/5 = 12x-3y/5

In ABD, cos d = 4x/5x = 4/5 -------- (3)

In NCD cos d = ND/4x-y ----------(4)
(3)=(4) thus:

4/5 = ND/4x-y.
ND = 4(4x-y)/5 = (16x-4y)/5 ---------(5)

AN = AD – ND = 5x – (16x –4y)/5 = (25x –16x + 4y)/5 = (9x + 4y)/5

QED

^^Kai!

You're right!

Foot of a perpendicular line should be displayed as Neo's diagram shows!

long yawn....

@OP,.....So You Really Want To Pass Your School Assignment Abii ?....I`ll Make Sure Your Lecture Know About This !!! Infact Do You Want To Scatter Our Head ?? YABC NDLEA OPC YSL Haba lol

@FACE, what is the difference between my solution and yours?

All this people we sabi book sef...*runs away from thread*

So let's first draw a picture:

@ saintneo, una too sabi book na why u no read wetin Faze write for dia ooo. The guy dey hail u fa, say of all d ppl wey calc d maths na ur own lecturer carry full mark give! All dis mathinairalanders sha!... runing away from d thread...