secant and tangent

J

jazziza87

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Apr 2, 2006
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I got a problem in Geometry that deals with tangents and secants.

My problem has one tangent and secant, the tangent is located at the top of the circle and the secant is a little bit down from the circle. I am asked to solve for x. The secant and the tangent intersect outside the circle and that intersection would be x degrees. Next they give me the inner arc which is 63 degrees and the outer arc is equal to 5x degrees.

I have to use the theorem that says that the measure of angle x equals one-half of the outer arc minus the inner arc. I am having trouble getting the outer arc. Could someone please help me. :?

Thanks in advance
 
What you are saying, is that, according to your theorem, x = ½(5x - 63º)?

Solve for 'x'. Multiply by 5 to get the measure of the outer arc.
 
sorry I don't understand what you are saying. Are you saying to multiply five by one-half? And if so it would come out to two and one-half what would I do then? Could you please explain a bit more
Thanks.
 
jazziza87, I answered or tried to answer several problems for you.
Please do us a favor and show us you can solve this:
\(\displaystyle x = \frac{{\left( {5x - 63} \right)}}{2}\) for x. Please do that.
 
Pka, if you don't want to help me that is fine, but you have to understand something here, I am homeschooled and don't have any tutor my parents have no clue on how to do my math and I know you are going to say go to school, but I recently moved and came to late in the year to enroll in school. So put yourself in my shoes, when you are trying to learn Geometry all by yourself. But I do thank you for your help, so if you don't want to help me out,don't!
Click to expand...
pka IS doing exactly as you request. As a professional educator, an important task is to assess where the student IS so that proper assistance can be given. The most important thing you could do is simply stop making excuses and start answering questions that will help others help you. You are studying math on your own. So? Does this qualify you to understand how best to teach mathematical concepts? Even to yourself? Absolutely not. If you want help, cooperation is the best way to go. soroban is very honest in pointing out that you provided no work whatsoever on this problem. Simply being shown how to do one problem WILL NOT provide for you a sufficient background for future problems. YOU must do the work.

That's enough of that. Let's learn some math.
 
Hello, jazziza87!

My problem has one tangent and secant.
The tangent is located at the top of the circle and the secant is a little bit down from the circle.
I am asked to solve for x.
The secant and the tangent intersect outside the circle and that intersection would be \(\displaystyle x\) degrees.
Next they give me the inner arc which is \(\displaystyle 63^o\) and the outer arc is equal to \(\displaystyle 5x\) degrees.

I have to use the theorem that says that the measure of angle x
equals one-half of the outer arc minus the inner arc.
I am having trouble getting the outer arc. . . . Really?
Click to expand...
Code: 
                B
              * * * - - - - - * A
          *           *   x /
        *           63° * /
       *                 * C
                        /
      *               /   *
      *             /     *
      * 5x        /       *
                /
       *      /          *
        *   /           *
          *           *
        D     * * *

The problem gave you the formula for the angle at \(\displaystyle A\)
\(\displaystyle \;\;\)and they gave you all the necessary parts.

So far, you haven't assured us that you even got this far:
\(\displaystyle \angle A\;=\;\frac{1}{2}(\text{outer arc}\,-\, \text{inner arc})\)
\(\displaystyle \;\;\downarrow\;\;\;\;\;\;\;\;\;\downarrow\;\;\;\;\;\;\;\;\downarrow\)
\(\displaystyle \:\:x\;\;\;=\;\frac{1}{2}\;(\;\,5x\;\;\;-\;\;\;63\;\;)\)
or that you even understand why we wrote it.
Then it sounds like you don't know enough algebra to solve the equation.

You want us to help . . . well, what have we been doing?
We've been pointing you toward the solution.
It requires some understanding (and cooperation) on your part.

If you want just the answer, it's \(\displaystyle 21^o\) . . . happy now?
 
could you show me how you got 21 degrees, you didn't have to give me the answer but thanks, just show me the steps you took to get the answer and I will be happy because than I can do future problems by myself.
 
Hello, jazziza87!

And you still haven't told us anything . . . oh, well . . .

Solving that equation won't help if you don't know where it came from.
That is 99% of the problem: knowing how to set up the equation.

After that, it's just nuts-and-bolts Algebra . . .
\(\displaystyle \;\;\)do this, do that and get the answer.


Okay, I'll take baby-steps . . .

We have: \(\displaystyle \,x\:=\:\frac{1}{2}\cdot(5x\,-\,63)\)

Multiply both sides by 2: \(\displaystyle \:2\cdot x\;= \;2\cdot\frac{1}{2}\cdot(5x\,-\,63)\)

And we have: \(\displaystyle \:2x\;=\;5x\,-\,63\)

Subtract \(\displaystyle 5x\) from both sides: \(\displaystyle \:2x\,-\,5x\;=\;5x\,-\,63\,-\,5x\)

And we have: \(\displaystyle \;-3x\;=\;-63\)

Divide both sides by -3: \(\displaystyle \;\frac{-3x}{-3}\;=\;\frac{-63}{-3}\)

Therefore: \(\displaystyle \;x\:=\:21\)
 
jazziza87 said:
could you show me how you got 21 degrees, just show me the steps you took to get the answer and I will be happy because than I can do future problems by myself.
Click to expand...
I find that highly doubtful! One needs to understand basic algebra before doing geometry. Maybe you should have a talk with your parents. Your homeschooling needs to address your weakness in algebra before moving you on into geometry.
Please have that talk with them.

\(\displaystyle \begin{array}{c}
x & = & \frac{{5x - 63}}{2} \\
2x & = & 5x - 63 \\
- 3x & = & - 63 \\
x & = & 21 \\
\end{array}\).
 
Thanks Soroban I did take Algebra 1 but I did forget some formulas thanks for refreshing my Algebra.To prove to you that I do remember now especially to pka who thinks that I don't know Algebra at all I will do another problem. Although I do think he is ignorant!

x=1\2(12x-32)
2x=12x-32
-10x=-32
x=3.2
 
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