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#1 2015-11-10 11:57:19

evene
Member
Registered: 2015-10-18
Posts: 272

Trigonometry!

Just need help with these "simple" Trigonometry questions!⇐ Meh...


(1)My eyes are 300 feet from the base of a very tall building. The building is on a slight hill, so that when I look straight ahead, I am staring at the base of the building. When I look upward at an angle of 54 degrees, I am looking at the top of the building. To the nearest foot, how many feet tall is the building?


(2)Degrees are not the only units we use to measure angles. We also use radians. Just as there are 360 in a circle, there are 2π radians in a circle. Compute tan π/3

(3)If A is an acute angle such that tan A + sec A = 2, then find cos A.



P.S: Can you link me the main Latex page on this website as well? I am interested in how it works! Thanks

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#2 2015-11-10 12:42:19

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

Re: Trigonometry!

Hi;

Always start with a diagram.

1)

iu3C2ZB.png

height of building = opposite -> 412.9145761413521

2) pi radians is 180 degrees, so pi / 3 radians is 60 degrees. Can you do the rest?


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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#3 2015-11-10 19:54:59

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

hi evene,

The latex tutorial is at http://www.mathisfunforum.com/viewtopic.php?id=4397

Q3.  That equation can be re-written with only cos.

Now use

So now the equation has only cos(A).  Subtract 1 from both sides so you can square to eliminate the square root sign.

Re-arrange and solve the quadratic.  You will find two values for cos(A).  One does not fit the original equation so can be discarded; the other does fit as you can easily show.

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2015-11-11 12:34:51

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

Ok thanks, I will let you know if I need any more help...


First attempt:

It worked!!! Awesome!

Last edited by evene (2016-03-28 14:50:57)

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#5 2015-11-13 10:55:04

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

Need help with this problem:


In triangle ABC,
and
. Let
be the angle bisector of
.

(a) Prove that
.

(b) Let

and let
. Using similar triangles ABC and BCD, write an equation relating x and y.

(c) Write the equation from Part b in terms of

and find r.

(d) Compute

and
using Parts a-c. (Do not use a calculator!)

Last edited by evene (2016-03-28 14:51:32)

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#6 2015-11-13 21:13:58

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

hi evene, this has been asked before, but it won't take a moment to write it again.  smile

(a) If BD is an angle bisector then ABD = DBC = 36.  Now you can write all the angles as either 36 or 72 so there are three isosceles triangles altogether.  The required result follows straight away.

(b)  The two triangles are 36-72-72 so they are similar.  I like to write the two sets of letters one above the other, with corresponding vertices above/below.  That makes it easy to write out the ratio equation.

ABC
BCD

So

Put in x, y and in one case x + y, to get an equation in x and y.

Use this:

So you can substitute with 'r' and get a quadratic in r.

Finally, in triangle ABC you can put in E, midpoint of BC, so that ABE is a right angled triangle with an angle of 72.  You can write the cos(72) in terms of x and y and hence get its value from the quadratic solutions.

A similar result for cos(36) can be found by putting in F, the midpoint of AB, to create another right angled triangle, BFD with an angle of 36.

Hope that's enough for you to complete this.  smile

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#7 2015-11-16 12:49:20

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

New Problems!

(a) A soccer ball is constructed using 32 regular polygons with equal side lengths. Twelve of the polygons are pentagons, and the rest are hexagons. A seam is sewn wherever two edges meet. What is the number of seams in the soccer ball?

What??


(b)Two circles of radius 1 are externally tangent at

. Let
and
be diameters of the two circles. From
a tangent is drawn to the circle with diameter
, and from
a parallel tangent is drawn to the circle with diameter
. Find the distance between these two tangent lines.


Need help with these! D:

Last edited by evene (2015-11-16 12:49:41)

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#8 2015-11-16 20:31:14

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

hi evene,

You are told there are 12 pentagons; so there must be 20 hexagons.

If they are separate bits you can work out how many edges there are altogether.  When it is sewn together  a pair of edges are joined so there must be half as many sewn edges as there are edges when apart.

I'm still looking at the circles problem.  I'm having trouble getting a sensible diagram.  Once I have that I'm hoping the rest will be easy.  dizzy Fingers crossed.

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#9 2015-11-16 21:26:38

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

Re: Trigonometry!

Hi


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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#10 2015-11-17 00:41:22

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

Hi bobbym,

I'm still having trouble with the diagram.  If I draw two random circles, overlapping an amount and fix the tangents, then the parallel property fails.  If I fix the parallel and one tangent, then the other isn't.  So any old circles won't do.  Hardly surprising in view of the question.  So I've tried moving points, trying to make all the properties work at once.  I've found one solution:  equal circles with PQR a straight line.  So I've been trying to prove that must be the case.  If PQR is straight then I can prove equal circles.  But I'm still stuck proving PQR is straight.

And then again, maybe this is just a special case anyway.

Is your answer that number multiplied by the radius ?  Surely any solution must be expressed in terms of the scale.

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#11 2015-11-17 01:37:26

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

How do we post diagrams on this website?

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#12 2015-11-17 04:42:02

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

hi evene,

The old server for this website had a small image upload feature, but when we switched to a new provider that was lost.  I have signed up to imgur.com and I upload images there.  You can then copy the BC code into your post and the image appears.

If you have a diagram, great!  If you cannot post it, you could just describe it

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#13 2015-11-17 04:43:05

phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,810

Re: Trigonometry!

Hi;

Here's my drawing of post#7(b)..sorry it's rather large:

vmepXDB.jpg

ΔPEB is 1/3rd of ΔPCA (similar right-angled triangles, with hypotenuse PB/PA = 1/3), and so EB = CA/3.

Therefore ED (the distance between the two tangent lines) = 4/3.

evene, the image is one I uploaded to Imgur (the file host I use), and if you click on the "Quote" button in the bottom right-hand corner of my post you'll see how I did it...by placing the image's url between the img tag.

Last edited by phrontister (2017-02-26 23:38:28)


"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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#14 2015-11-17 05:13:55

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

Re: Trigonometry!

Hi Bob;

The circles are externally tangent so take a look at phrontister's drawing. I gave the distance from his points D and C which is how I interpreted the question. I now see that his ( phrontister ) view is the correct one.


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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#15 2015-11-17 05:20:03

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

Did it always say:

Two circles of radius 1 are externally tangent at Q.

I've been trying to do it without this.  eek

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#16 2015-11-17 05:26:51

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

Re: Trigonometry!

Hi;

There was an edit to the post...


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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#17 2015-11-17 10:20:30

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

phrontister's diagram is what it is supposed to look like! Great job!

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#18 2015-11-17 14:45:58

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

Still need help with post 4. Only the last part though. I still need help computing


I have drawn

and see that
But that is all I see

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#19 2015-11-17 20:04:00

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

hi evene

In triangle ABE, angle ABE = 72 and BEA = 90 so we can use cos(72) = BE/AB = 0.5x/(x+y)

Divide top and bottom by x .... cos(72) = 0.5/(1+r)

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#20 2015-11-19 13:04:36

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

Ok thanks, I managed to solve the problem!

The circle problem. I put in 2, and the answer was incorrect. hmm

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#21 2015-11-19 20:20:11

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

hi evene,

2 ? Where did that come from ?

I thought phrontister had a good solution in post 13.

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#22 2015-11-20 01:39:55

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

oh...

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#23 2015-11-20 01:43:02

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

New problem

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
is
.

Last edited by evene (2015-11-20 01:43:33)

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#24 2015-11-20 06:37:16

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Trigonometry!

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
is
.

I assume that AB is not the tangent through P.

Once again I find I cannot make a diagram because there is insufficient information.  Are A and B the points on the two circles where the common tangent touches ?  And where are C and D ?

Please clarify, or better still, post a diagram.  Thanks,

Later Edit:  Is this the problem?

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

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#25 2015-11-22 03:39:22

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Trigonometry!

Oh well... yes and I see that there is another post about this problem too...

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