+0

# /gone

-3
141
5

/gone

Dec 14, 2019
edited by whoisjoe  Jan 7, 2020

#1
-1

(1) So simple!  Minor TU is 180 - 32 = 148 degrees.

(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.

Dec 14, 2019
#4
0

The answer for number 3 is not 11

Guest Dec 17, 2019
#2
+1

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   Dec 14, 2019
#3
+1

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   Dec 14, 2019
#5
0

For #3 you can just set up ratios.

Let PC = x.

Now you can set up the ratio.

(x/(x+8)=3/5

now you just solve the ratio and the answer should be 12.

Dec 17, 2019