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Two tangents to a circle with center $O$ meet at $A$ and extent an angle of $50^\circ$. Another tangent to the circle intersects the two tangents at $B$ and $C$. If $\angle BOC = x^\circ$, find $x$.

 

 Jan 26, 2021
 #1
avatar+1028 
+1

x = 65 degrees

 Jan 26, 2021
 #2
avatar+112523 
0

How did you get that Jugoslav?

Melody  Jan 30, 2021
 #3
avatar+112523 
+2

If you take the extreme case where BC and BA are the same tangent.

 

then angle ABO is 90 degrees, angle BAO=25 degrees  so   angle x must be  180-25-90 = 65 degrees.

 

There is obviously some theory that Jugoslav knows that says this will always be the case.

 

I don't know how to prove the general case.

 Jan 30, 2021
 #4
avatar+186 
+3

Extend AB to B' and AC to C'.

Call angle CBO alpha, then angle OBB' = alpha, so angle ABC = 180 - 2(alpha).

Call angle BCO beta, then angle ACB = 180 - 2(beta).

From triangle ABC, (alpha + beta) = 115 deg, and then from triangle BOC, x = 65 deg.

 Jan 31, 2021
 #5
avatar+2155 
+1

Wonderful Tiggsy!

Except, without a bloody diagram, I’m lost to where B’ and C’ are supposed to be, relative to the circle.  Are B’ and C’ positioned such that when they are joined by a vertical line, the line forms a tangent on the opposite side of the circle, mirroring the tangent of the line BC?

 

 

GA

GingerAle  Jan 31, 2021
edited by GingerAle  Jan 31, 2021
 #6
avatar+116125 
+1

Good work, Tiggsy......could   you provide a sketch   of  what you did   ???

 

 

 

cool cool cool

CPhill  Jan 31, 2021
 #7
avatar+31703 
+3

I suspect Tiggsy meant the following for B' (with an analogous point for C'):

 Jan 31, 2021
 #8
avatar+112523 
+3

Thanks Tiggsy,

 

I still have a question,

 

How do you know that   

 

angle CBO = angle OBB'      ?

 

Here is the pic to go with Tiggsy's explanation.

 

It is interactive, you can grab the point T and move it around to see that the result says the same.

 

https://www.geogebra.org/classic/hqcjnaku

 

 

Here is the non interactive version:

 

 Jan 31, 2021
edited by Melody  Jan 31, 2021
 #9
avatar+116125 
+1

.....

I still have a question,

 

How do you know that   

 

angle CBO = angle OBB'      ?....

 

 

 

 

Yeah.....I  was wondering about that too....

 

 

 

cool cool cool

CPhill  Jan 31, 2021
 #10
avatar+112523 
+1

I get it Chris, it is because BB'  and BC are both tangents and BO goes through the centre of the circle, so

 

angle  CBO and angle OBB' have to be congruent.

 

You could use the 3 equal sides test of congruency. 

Plus if two tangents intersect then the distances from the cross point to the circle intersections will be congruent.

Melody  Jan 31, 2021
edited by Melody  Jan 31, 2021
edited by Melody  Jan 31, 2021
 #11
avatar+116125 
+1

OK, got it   (duh!!!!)

 

Thanks, Melody   !!!!

 

cool cool cool

CPhill  Jan 31, 2021

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