I don\'t follow your thinking
If the do, it is an easy problem.
draw a line from the vertex through the two centres.
Let the distance between the vertex and the centre of the smaller circle be x
then the distance from the vertex to the centre of the larger circle is x+6
sin (theta/2) = 2/x
sin (theta/2) = 4/(x+6)
then 2/x = 4/(x+6)
which solves for x=6
then sin (theta/2) = 2/6
theta/2 = 19.47º
theta = 38.942º or 38.9º to the nearest tenth.
3 answers
circles do touch
what is this half angle identity thingy
sine theta/2 is equal to 2/x
why please explain
sine theta/2 is equal to 2/x
why please explain
the circles did not touch in your diagram.
In your diagram, theta was the entire angle between the two langent lines.
So the line to the centre would bisect the angle, in other words, theta/2.
Can you not see that we now created right-angles triangles?
I then just used the primary trig relationship, sin(angle) = opposite/hypotenuse
thus, sin(theta/2) = 2/x etc
In your diagram, theta was the entire angle between the two langent lines.
So the line to the centre would bisect the angle, in other words, theta/2.
Can you not see that we now created right-angles triangles?
I then just used the primary trig relationship, sin(angle) = opposite/hypotenuse
thus, sin(theta/2) = 2/x etc
0 answers