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#1 2013-07-31 06:48:44

cooljackiec
Member
Registered: 2012-12-13
Posts: 160

trig

I'm standing at 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?

What is the cosine of the angle between two adjacent faces of a regular tetrahedron?


I see you have graph paper.
You must be plotting something
lol

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#2 2013-07-31 07:22:05

atran
Member
Registered: 2013-07-12
Posts: 91

Re: trig

tan(54⁰) = height / 300

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#3 2013-08-06 10:27:09

cooljackiec
Member
Registered: 2012-12-13
Posts: 160

Re: trig

any feedback on the 2nd problem?

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


I see you have graph paper.
You must be plotting something
lol

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#4 2013-08-06 11:06:37

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,308

Re: trig

http://www.mathisfunforum.com/viewtopic … 98#p196898

use tan = sin/cos and sec = 1/cos

Re-write the equation in the form  R(cosBcosA - sinBsinA) where B can be determined.

Hence get cos(A+B)

EDIT:  better to try this

Write as sinA = function of cosA

Square and form a quadratic in cos using sin^2 = 1 - cos^2


Bob


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

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#5 2013-08-07 09:36:17

cooljackiec
Member
Registered: 2012-12-13
Posts: 160

Re: trig

i am not sure what you mean for the dihedral angle problem.


I see you have graph paper.
You must be plotting something
lol

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#6 2013-08-07 20:00:06

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,308

Re: trig

hi cooljackiec,

In 2 dimensions, when you look at an angle it is obvious what size it is.  This is more difficult in 3 dimensions.  Here's an experiment you can try that will show what I mean:

Cut a triangle ABC out of a sheet of paper.  If you can,  make it fairly large and make angle A obtuse.  Of course, B and C will be acute.

Now go to the corner of your room and put A into the corner.  Juggle with the position of B and C until AB lies against one wall and AC lies against the other.  (hint: Out of B and C,  one will need to be lower down the wall than A and the other higher.)

Now try to fit B into the corner. (This time both A and C will need to be lower than B, with BA touching one wall and BC touching the other.)

This will work whatever the size of the angles.  So what is the true angle between the walls?  By choosing A at the corner, B in one wall and C in the other, I have shown that any obtuse angle will 'fit'.  With B at the corner, I have shown that any acute angle will fit.

But, you know that the angle between the walls is actually 90.  So how can we 'see' this true angle and measure the angle between faces correctly.

Look at my diagram below.  You will 'see' the angle between the walls as 90 if you are looking along the dotted line.  When you look in the direction of the arrow, the angle looks like it is 90 as it should.

To do the same for the angle between faces on a tetrahedron you need to imagine you are looking along the line of intersection between two faces.

In that other thread for faces VAB and VBC, the line to look along is VB.  The true angle between the faces is the size of angle AGC, where G is on VB and both AG and CG are at right angles to VB.

The second diagram in that thread (post 3) shows the true shape of face VBC.  So my method is to work out CG from that.  For a regular tetrahedron the angle B is 60.  AG will be the same length as CG.

Then you can use the cosine rule to calculate AGC.

Bob

View Image: cooljackiec12.gif

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

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