Help finding an angle

Jamz

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Apr 20, 2013
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I've been trying to figure out this question for quite some time now.. and I have no idea how to find the answer. If I had a single angle it would be much easier!




Thanks in advance!:D
 
Hello, Jamz!

I've been trying to figure out this question for quite some time now.
I have no idea how to find the answer. .If I had a single angle it would be much easier!


We need more information.

I will assume that the circular arc is exactly a quarter-circle.
Code:
                    * * A
                *     |
              *       |
             *        | 1
                      |
      F     *         |
      :   B o - - - - * O
      :    /:    1    |      
      :   / :         |
      :θ /  :         | 0.863
      : /   :         |
      :/ α  :         |
    C * - - * - - - - * E
       0.25 D    1    :
      : - -  1.25 - - :
The center of the circle is O.  AOB=90o\displaystyle O.\;\angle AOB = 90^o
. . AO=BO=DE=1\displaystyle AO = BO \,=\, DE \,=\, 1

We find that: .OE=BD=0.863\displaystyle OE = BD = 0.863
We find that: .CE=1.25CD=0.25\displaystyle CE = 1.25 \quad\Rightarrow\quad CD = 0.25

Let α=BCD,  θ=BCF\displaystyle \alpha = \angle BCD,\;\theta = \angle BCF

In ΔBDC ⁣:  tanα=BDCD=0.8630.25=3.452\displaystyle \Delta BDC\!:\;\tan\alpha \,=\,\dfrac{BD}{CD} \,=\,\dfrac{0.863}{0.25} \:=\:3.452

Hence: .α=74.80437745o73.8o\displaystyle \alpha \:=\:74.80437745^o \:\approx\:73.8^o

Therefore: .θ=90o73.8o  =  16.2o\displaystyle \theta \:=\:90^o - 73.8^o \;=\;16.2^o
 
I've been trying to figure out this question for quite some time now.. and I have no idea how to find the answer. If I had a single angle it would be much easier!




Thanks in advance!:D
The sloping line is tangent to the circle, and passes through a known point. Lets identify that as point A, which is 2.062 from the left edge and 0.500 above the base. The math will be much easier if we shift the origin of the coordinate system to be at the center of the circle, which I make to be (2.873-0.510-1.000) = 1.363 above the base. After shifting the origin, the equation of the circle is
x^2 + y^2 = 1.000
and point A is
A = (-1.250, -0.863)

Do you know how to find the equation of the line through point A that is tangent to the circle?
 
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