Asked by Trigonemtry REINY

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.
Answered by Trigonemtry REINY
circles do touch
Answered by Trigonemtry REINY
what is this half angle identity thingy

sine theta/2 is equal to 2/x

why please explain
Answered by Reiny
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
There are no AI answers yet. The ability to request AI answers is coming soon!