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**championmathgirl****Member**- Registered: 2015-06-01
- Posts: 20

In triangle ABC, angle A = 36 degrees and angle B = angle C = 72 degrees. Let BD be the angle bisector of angle ABC.

(a) Prove that BC = BD = AD.

(b) Let x = BC and let y = CD. Using similar triangles ABC and BCD, write an equation relating x and y.

(c) Write the equation from Part (b) in terms of r=y/x and find r.

(d) Compute cos 36 degrees and cos 72 degrees using Parts a-c. (Do not use a calculator!)

I got part (a), you can easily find and use the missing angles to prove BC = BD = AD. But I need a bit of help with the other parts.

Thanks!

Girls can be just as good as boys at math. We just need to get the same encouragement.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

hi championmathgirl

BC = BD = AD = x

So AC = AB = x + y

Using the similar triangles

Divide all by x^2 and replace y/x by r, leading to a quadratic in r, which you will need to solve using the formula (or completing the square).

Two solutions, but y/x is obviously positive so you can reject the negative solution.

You can use the cosine rule or split a 36, 72, 72 triangle along its line of symmetry to get cos36 and cos72, firstly in terms of x and y, and then in terms of r, to finish the question.

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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**championmathgirl****Member**- Registered: 2015-06-01
- Posts: 20

Erm... I'm kinda stuck. I got to part (c) and well I got the equation r^2+xr-1=0. Even using the quadratic or completing the square, i get stuck with a square root with a lot of variables.

I want to check to make sure the equation is right, and what i got for r is also right. Then I need a bit of help with the rest.

Super thanks!!

Girls can be just as good as boys at math. We just need to get the same encouragement.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Using Bob's last equation:

Now using the first equation:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

Alternatively you could do this

Either way you shouldn't have an 'x' left in the 'r' quadratic.

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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**championmathgirl****Member**- Registered: 2015-06-01
- Posts: 20

Is r=(1+sqrt3)/2? And is Cos 36 = 1/2+1/2*r, Cos 72=1/(2+2r)

I checked it, and it didn't seem to work out. Where is it that I went wrong?

*Last edited by championmathgirl (2015-08-05 12:55:26)*

Girls can be just as good as boys at math. We just need to get the same encouragement.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

It looks like you have a sign error in the quadratic formula.

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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**Shrimpy****Member**- Registered: 2017-02-03
- Posts: 4

So...I finished everything up to part c. I can find $\cos 72$ by using triangle BCD, by splitting it in half, but how do I find $\cos 36$?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

hi Shrimpy

Welcome to the forum.

It's a while since I did this question but I seems to remember that you can get cos36 from another right angled triangle in the diagram. Make a sketch and split a 36-36-108 triangle down the middle.

If you're still not getting there post back and I'll start the question again.

Bob

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

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**Geomtry.202****Member**- Registered: 2017-03-13
- Posts: 2

I need help in trying to find cos(36). How will splitting the 36-36-108 help?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

hi Geomtry.202

Welcome to the forum.

When you split an isosceles triangle along its line of symmetry, you create a right angled triangle. So you can determine cos(36) using A/H.

Bob

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

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Can you walk me through parts (b), (c), and (d)? I'm having a little trouble with them. (a) is easy enough, though.

Many thanks!

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

hi !nval!d_us3rnam3

(b) Let x = BC and let y = CD. Using similar triangles ABC and BCD, write an equation relating x and y.

(c) Write the equation from Part (b) in terms of r=y/x and find r.

(d) Compute cos 36 degrees and cos 72 degrees using Parts a-c.

So identify two isosceles triangles, with 36-72-72 angles.

Write the letters for these, in order, one below the other. This makes it easier to construct the ratios you'll need.

Fill in every length on the diagram with x and y as appropriate.

Now write the ratio r = x/y in all ways possible using the similar triangles.

From these you should be able to eliminate x and y and so get an equation for r.

If you split an isosceles triangle down its line of symmetry you'll get right angled triangles with 36 and 72 in them, so you can then use A/H to get the cosines, initially with x and y, but then convert to 'r' and use the known value from above. You can check your answer by evaluating it and seeing if your calculator agrees using the cos button.

Bob

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

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Can you actually walk me through part (d)? I did (a), (b), and (c).

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

let's check you have my answers to those first. Please post.

Bob

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

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Yknow, I did it already, but I'll still post my answers for all of them.

(b) x^2=y^2+xy

(c)r^2+r=1, r = {sqrt(5)-1}/2

(d) cos 36 = {sqrt(5)+1}/4, cos 72 = {sqrt(5)-1}/4

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

hi !nval!d_us3rnam3

That's great! Glad you have got it sorted. Funny thing though. I've got some sign differences in the (c) answers. Same final result.

Bob

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

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

What are the sign differences, may I ask? Just wanna make sure I didn't miss something.

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

Sorry, my mistake. I should have worked with r = y/x, but I wrote r = x/y instead. So I got a similar quadratic and value for r, leading to the same cos(36) and cos(72). But not the same quadratic as you. That explains for me why your (d) answers were also mine. In re-assessing your results, I now think you are completely correct. Apologies.

Bob

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

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Thank you!

I just need help with one more problem. (Please don't just give me a link to somewhere else, work me through some of it here.)

Two circles are externally tangent at point

, as shown. Segment is parallel to common external tangent Prove that the distance between the midpoints of and isEDIT: I extended lines AC and BD to meet at Q, then proved that QAB is similar to QCD. Then what?

And y duz the

*Last edited by !nval!d_us3rnam3 (2017-05-17 01:37:40)*

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,399

hi !nval!d_us3rnam3

(Please don't just give me a link to somewhere else, work me through some of it here.)

I'm sorry, but I think that request is unreasonable. This question has been dealt with before and it took a lot of time to put together all the help in this post:

http://www.mathisfunforum.com/viewtopic.php?id=22435

Please take the time to read it and post back if you meet something that is not clear to you.

Bob

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

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Em, whatever. I've finished it already

Thanks anyway!

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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