CPhill

                    A

                       2HC

        4               H        4

          B                   C

3HC =  4

HC = 4/3

AH = 8/3

BH^2  + AH^2  = AB^2

BH^2    (8/3)^2  = 4^2

BH^2  = 16 - 64/9  =  [144 - 64 ] / 9    =   80  / 9

BH^2 + HC^2  = BC^2

80/9 + 16/9  = BC^2

96/9 = BC^2

sqrt (96) / 3  =   (4/3)sqrt (6)  = BC

https://web2.0calc.com/questions/circles_146

A

      2

         H

C      1         B

Triangle AHC  ≈ Triangle  ACB

AH / CH  = AC / BC

2 / CH  = AC 

CH  = 2 / AC

Triangle BHC ≈ BCA

CH / BC  = AC / BA

CH  =  AC / BA

2 /AC  = AC / BA

2BA = AC^2

2BA + BC^2  = AB^2

2BA + 1^2 = AB^2

AB^2 - 2AB - 1  =  0

AB^2 - 2AB  =  1

AB^2  -2AB + 1 = 2

(AB - 1)^2  =  2

AB - 1  = sqrt(2)

AB = 1 + sqrt (2)

BH  =  AB  - AH   =  1  + sqrt (2)  - 2  =  sqrt (2) - 1

CH^2 + BH^2 = BC^2

CH^2 + (sqrt (2) - 1)^2  = 1^2

CH^2  + (1 - 2sqrt 2 + 2)  = 1

CH^2  = 2sqrt(2)  - 2

CH = sqrt [ 2sqrt (2) - 2 ] ≈  .91

2 / ( 4 + 1 + 2 + 3)  *  20   =  

2/ 10  * 20  =

40 / 10  =    4  blue

Angle B  =  180 -  55 -  26  =  99

Law of Sines

BC /  sin 55   =  10  / sin 99

BC =  10sin 55  / sin 99  ≈   8.29

The similar triangle will have sides of  20  - 20  - 16

AB / BC  =  XY   / YZ

14  / 8  =  16  / YZ

YZ / 16  =  8  / 14

YZ =  (16*8)  /14  =    (8/7) (8)  =  64  /  7

Let the center of  the small circle =  C

Let the tangent point of  the circle with OA = D

Triangle  OCD  is  right such that

OD^2  + DC^2  = OC^2

3^2  + r^2  = (5 - r)^2

9 + r^2  = r^2  - 10r  + 25

10r  = 16

r =  16 / 10   =  1.6

Just a slight mistake  by asinus

Slope of AD =  -1/3

We need to  solve this to find the x coordinate  of the  missing  vertex

(-2/5)(x + 3)  + 1   = (-1/3)(x -5) - 2

-6 ( x + 3) + 15  = -5 (x - 5)  - 30

-6x - 3  = -5x -5

x = 2

y =  (-2/5)(2 + 3) + 1   =  -1

The 4th vertex  =  (2. -1)     { This is in the 4th Q ,  not the 3rd Q }

Sum of  coordinates =  1

The two angles sum to 120

So.....the measure of the smallest angle is

5/ (5 + 7)  * 120    =

5 / 12  *   120   =

60°