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Two circles, whose radii are 12 and 16 inches respectively intersect. The angle between the tangents at either of the points of intersection is 29°30`. Find the distance between the centers of the circles.

If possible please show illustration. For further help, what do you mean by 'the angle between the tangents at either of the points of intersection' because this is where I am having a hard time not knowing where to put the 29 degrees 30 minutes angle. Thanks!

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up vote 1 down vote accepted

enter image description here

By Cosine rule in triangle, $$\cos\theta=\frac{r_1^2+r_2^2-d^2}{2r_1r_2}$$

You know, $r_1=12,r_2=16$ and $\theta=29^o30'=29.5^o$

Plug in these values, and evaluate $d$.

EDIT: $d^2=r_1^2+r_2^2-2r_1r_2\cos\theta=256+144-384\cos(29.5^o)=400-334.21approx 65.8\implies d\approx\sqrt{65.8}\approx 8.11$ inches

For you second problem in comment, enter image description here

You can apply the same formula $\cos\theta=\frac{15^2+7^2-20^2}{2(15)(7)}$

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how can i find d? what is the solution? – Ziii Jan 24 '13 at 10:30
but 8.11 is the whole length of the other side of triangle and we are looking for the distance between the centers of the circle. – Ziii Jan 24 '13 at 10:37
Now, i see your problem. Actually, tangent at point of contact passes through centers of the circles.The side of triangle with length $r_1$ actually represents the line joining point of contact and center of first circle. similar result follows for the other circle. So, $d$ is the distance between center of two circles. – Aang Jan 24 '13 at 10:45
Ohh! THANK YOU SO MUCHHH! :) – Ziii Jan 24 '13 at 11:05
Please help on this one too. – Ziii Jan 24 '13 at 11:06

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