ElectricPavlov

His resultant speed is    sqrt ( 1^2 + 2^2) = sqrt 5

he needs to swim at an angle....the distance he swims is the hypotenuse of a right triangle  and =   sqrt 5 t  

    where t is the time it takes him to cross....meanwhile the current moves him  1 km/hr * t

sin  = opp/hyp  =    1 t / (sqrt 5 * t) = 1/5

arcsin 1/5 = 26.56 degrees angled upstream       an angle of   90 - 26.56 =  63.43 degrees with the bank

Answered a question like this the other day....plot all of the given points and come back with your graph for more help....

Normal speed = x

xt = 200      t = 200/x

faster speed    (x+10 )    new time = t-1   = 200/x -1

(x+10)(200/x -1) = 200 miles

200 -x + 2000/x -10 = 200

-x  + 2000/x -10 = 0

-x^2 + 2000 - 10 x = 0       via quadratic formula   x = 40 m/hr

See original answer above...I edited it .....   

https://web2.0calc.com/questions/please-help-must-get-dinner

a = k/b^3      sub in a = 9/2    and b = 2   to find k

9/2 = k / 8

k = 36

a = 36/b^3   

   = 36/3³ = 36/27 = 1  1/3

cosx = -4/9     in quadrant 2 and 4

arccos (-4/9) = 2.03 R   and 4.25 R

(a + b + c )/3 = 9

c-4 = a    c = a+4

b+3       = 2/3 ( a +3)        b = 2/3( a+3)  - 3       Sub the red in to the top equation

(a + 2/3(a+3)-3   +   a+4 )/3 = 9

2  2/3 a   +3 = 27

8/3 a = 24

a = 9               then   c=a+4 = 13  y/o                 and b=5

**** deleted ****

csc is   1/sin    in quadrant III   and IV   sin is negative, so there should be two answers'

1/sin = -2 sqrt3/3

sin = 1/(-2 sqrt3/3)

arcsin = -60 degrees    and   240 degrees       or     240 and 300 degrees   or  4pi/3    and  5pi/3     this is   3x

divide the answers by 3     4pi/9   5pi/9      (edited)