Trig Problem

BobbyOLeary

Check it out, its a tough enough question though if you're doing honours you should be well able for it.

Looking at the attachment:

In the figure below, with lengths as labeled, AC = BC, CF = DF, the angle DBC and the angle EAC are right, and angle DFC is 30°. What is the length CF?

Delphi91

Right, use Pythagoras' Theorem on Triangle AEC. That gives |AC| = 12

Triangle ABC is isoceles (as |AC| = |BC|) therefore |BC| is also 12.

Use Pythagoras' Theorem on Triangle DBC. This gives |DC| = 15.

Triangle DCF is also isoceles (as |CF| = |DF|). This allows you to work out the angles DCF and CDF. They are both 75 degrees [(180-30)/2].

Now use the sine rule to give you: (Sin 30)/15 = (Sin75)/|CF|.

This gives |CF| = 30.


Hope this helps?

Mike

BobbyOLeary

Spot on.

I'm not actually doing the leaving, I'm in College at the moment. I was just posting up the problem because I thought it was a nice one. It was from some Trigonometry Contest a while back. It seems to be about Honours level, what level are you currently at Mike?

-Bobby

Delphi91

BobbyOLeary wrote:

It seems to be about Honours level, what level are you currently at Mike?
-Bobby

Well, I'm more in front of the desk than behind it, if you know what I mean!:D

To be honest, I'd be surprised if it was honours level - there really isn't anything too complicated that an ordinary level student couldn't figure out - basic enough trigonometry really.

ZorbaTehZ

How about this one:

In the triangle, |AP| = 5 and |BP| = 3.
Moreover, CP is the bisector of the angle <ACB.
Calculate Sin A/Sin B

Delphi91

ZorbaTehZ wrote:

How about this one:

In the triangle, |AP| = 5 and |BP| = 3.
Moreover, CP is the bisector of the angle <ACB.
Calculate Sin A/Sin B

(Sin A)/|CP| = (Sin [C/2])/5

and

(Sin /|CP| = (Sin [C/2])/3

Therefore cross-multiplying each of the above gives:

5(Sin A) = |CP|Sin[C/2]
and
3(Sin = |CP|Sin[C/2]

Therefore 5(Sin A) = 3(Sin

That gives Sin A/Sin B = 3/5.

ZorbaTehZ

Delphi91 wrote:

(Sin A)/|CP| = (Sin [C/2])/5

and

(Sin /|CP| = (Sin [C/2])/3

Therefore cross-multiplying each of the above gives:

5(Sin A) = |CP|Sin[C/2]
and
3(Sin = |CP|Sin[C/2]

Therefore 5(Sin A) = 3(Sin

That gives Sin A/Sin B = 3/5.

Nice one.

One more: (Will try make a pic.)

Two circles S and T intersect at distinct points A and B. Suppose that L is the line containing A and B and suppose that M is a line that is a tangent to S at P and Tangent to T at R. Show that L meets M at the midpoint of P and R.
(Remember S and T are not necessarily of equal radius).

BTW Delphi, teacher/student?

EDIT: Included the pic.
Its not that good - used paint - but should give a general idea.

Delphi91

ZorbaTehZ wrote:

Nice one.

One more: (Will try make a pic.)

Two circles S and T intersect at distinct points A and B. Suppose that L is the line containing A and B and suppose that M is a line that is a tangent to S at P and Tangent to T at R. Show that L meets M at the midpoint of P and R.
(Remember S and T are not necessarily of equal radius).

BTW Delphi, teacher/student?

Where are P and R??

And I'm a teacher.