CPhill

A =  what Alice has at first

B = what Bob has at first

(A + n)  = 10(B - n)   →  A + n = 10B  - 10n  →   A  -10B  =  -11n     (1)

(A - n)  = 4 (B + n)   → A - n  =  4B + 4n       →   A - 4B  =  5n     (2)

Multiply  (1)  by 5   and (2)  by 11

5A - 50B  = -55n

11A - 44B  = 55n          add these

16A - 94B =  0

16A  = 94B

A / B  =  94 / 16 =  47 /  8

sin (ABC)   =sin (90 degrees)  =  1

tan x + sec x =  2cot x

(sin x) / (cos x) + 1 /cos x)  =   2 (cos x)/(sin x)

( sin x + 1)  / cos x  =   2 (cos x) / (sin x)

(sin x + 1) / cos x   -  2 cos x / sin x  =   0

[sin x  (sinx + 1)  - 2cos ^2 x ] / [ sin x cos x]  = 0

sin ^2 x + sin x - 2cos^2 x  =  0

sin^2 x + sin x - 2 ( 1 - sin^2 x)  =  0

3sin^2 x + sin x - 2  =  0      

(3sin x - 2) (sin x  + 1)  = 0

3sin x - 2  = 0                   sinx + 1  = 0

3sinx  = 2                        sinx  = -1

 sin x = 2/3                       x = -3pi/2        (reject)

cos x  =  sqrt  ( 1 -(2/3)^2 )  =  sqrt  (1 - 4/9)  =  sqrt (5) /  3   

The denominator cannot  = 0

Therefore, every real number except x = 7  is in the domain

So

(-inf, 7) U ( 7 , inf)

The exterior angles sum to  360°

360 -  120  -  75   =    165°

Simplify  as

6x^2 + ( k - 12)x + 9  = 0

We have non-real roots when the discriminant > 0

So

(k  -12)^2  - 4(6)(9) < 0

(k -12)^2  -  216 <  0

(k -12)^2  < 216    take both roots

k - 12 < sqrt (216)                and        k - 12 >  -sqrt (216)

k < 12 + 6sqrt (6)                                 k > 12 - 6 sqrt (6) 

k  on this interval →  ( 12 -6sqrt(6)  , 12 + 6sqrt (6) )                                     

-b =  (4 + 2i)  + (-5 + 8i)  =  -1 + 10i

b = (1 -10i)

c = (4 + 2i) (-5 + 8i)  =   -20 -10i + 32i + 16i^2  =  -36 +22i

b + c  =  (-1 + 10i)  + ( -36 + 22i)  =   -37 + 32i

3^(2c + 1)  = 28 *3^c  + 4*3^c -25 - 9

3* 3^(2c)  = 32*3^c  - 34

3* (3^c)^2  -  32*3^c  + 34  = 0          let  3^c  = x     so  that  (3^c)^2  = x^2

So we have

3x^2  - 32x +  34 =  0

Using the Quadratic Formula

x =    [ 32 + sqrt [ 32^2 - 4 * 3 *34] ]  / 6    =    [ 32 + sqrt (616) ] / 6  = 

[32 +2sqrt (154) ] / 6  =  [16 + sqrt (154) ] / 3

And x = the conjugate  =  [ 16 -sqrt (154) ] / 3

So

3^c  = [ 16 - sqrt (154) ] / 3       take the log of  both  sides

log 3^c  = log ( [16 - sqrt (154)] / 3)

c log 3  = log ( [ 16 - sqrt (154)]/3)

c = log ( [16 -sqrt (154)]/3) / log 3 ≈ .163         and   

c  =  log ([16 + sqrt (154) ] / 3) / log 3  ≈  2.046

Something like this ( Not to Scale)

 By the property of medians  ON = (1/3)QN = (1/3)(1) = 1/3

And  since N is  a midpoint of PR then  NR = (1/2)PR =  1/2

And QN perp to PR  so   triangle RNO is right with angle RNO  =  90

By the Pythagorean Theorem

OR = sqrt (ON^2  + NR^2)   = sqrt [ (1/3)^2  + (1/2)^2 ] =  sqrt [ 1/9 + 1/4]  = sqrt  [ 13 / 36 ]  = 

sqrt (13) / 6  ≈  0.6

We need to know BC