Puzzling Trig Problem Need Help.

  • Thread starter armolinasf
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  • #1
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Homework Statement



I came across this interesting Trig problem that involves an angle θ formed by two rays OB and OA. there are two circles, one with radius a and one with radius b, on on the ray OA that are both tangent to one another. The ray OB is tangent to both of the circles.

Show that cos(θ)=(ab)^(1/2)/(a+b)/2, in other words that cosine is equal to the ratio of the geometric mean to the arithmetic mean.

Homework Equations



Sin(θ)=(b-a)/(b+a)



The Attempt at a Solution



I understand that going from sin to cos is just a matter of applying the pythagorean identity, what I'm having trouble understanding is why sin(θ)= (b-a)/(b+a).

Thanks for the help.
 

Answers and Replies

  • #2
Dick
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Draw a right triangle. The hypotenuse is the segment connecting the centers of the two circles. Make one of the legs parallel to one of the rays and the other perpendicular. Do you see it? And I think the angle theta is half of the angle between the rays, isn't it?
 
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  • #3
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Alright, so now I have a right triangle with a hypotenuse b+a, it's similar to the right triangle that could have been formed by the angle theta and the radius b of the larger circle. The angle of the small triangle should be equal to theta since its two parallel lines crossed by a transversal (unless I misinterpreted you).

Hypotenuse=b+a makes perfect sense, i can see it. But the side opposite the angle theta of the smaller triangle is some length equal to b-x, how do we know that x=a?
 
  • #4
Dick
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Alright, so now I have a right triangle with a hypotenuse b+a, it's similar to the right triangle that could have been formed by the angle theta and the radius b of the larger circle. The angle of the small triangle should be equal to theta since its two parallel lines crossed by a transversal (unless I misinterpreted you).

Hypotenuse=b+a makes perfect sense, i can see it. But the side opposite the angle theta of the smaller triangle is some length equal to b-x, how do we know that x=a?

Drop perpendiculars from the ends of the hypotenuse to the ray. One end is distance a from the ray, the other end is distance b. The side opposite is the difference of those.
 
  • #5
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Got it and it makes sense, thanks for the help.
 

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