(1) So simple! Minor TU is 180 - 32 = 148 degrees.
(2) I will have to think about this one...
(3) The external tangent gives us similar triangles OTC and PUC. Let x = PC. Then x/(x + 5) = 3/8, so x = 3. The answer is PC = 3 + 3 + 5 = 11.
1. Points T and U are on a circle centered at O, and point P is outside the circle such that PT and PU are tangent to the circle. If angle OPT = 32 degrees, then what is the measure of minor arc TU, in degrees?
P
T U
O
Since PT is tangent tothe circle, angle OTP = 90°
And since OPT = 32¯, then angle POT = 180 - 90 - 32 = 58°
So....by symmetry, angle POU must also = 58°.....so central angle TOU = 2 (58) = 116° which will also be the measure of minor arc TU
Second one.....let A be the apex of the triangle....Let AB be the tangent drawn to the top circle tangent to this circle at B....and let C be the center of the top circle
So triangle ABC is a 30- 60 - 90 right triangle with angle CBA = 90°
CB = the radius of the top circle = 3
Angle CAB = (1/2)60° = 30°.....so angle ACB = 60°
So AB = CB*sqrt (3) = 3sqrt(3)
So....by symmetry.....the perimeter of the equilateral triangle = 3 times twice the radius of each circle + 6 * 3sqrt(3)
= 3 * 2 (3) + 18sqrt(3) =
18 + 18sqrt(3) =
18 [ 1 + sqrt(3) ] units