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Correct, history !!!

I haven't even learnt arctan yet :( I'm still learning tan. 

When sides line AB and line EF are extended, they form a straight line, which is a 180 degree angle. So, angle P is equal to 180 degrees - 120 degrees, which simplifies to 60 degrees. Therefore, the degree measure of angle P is 60 degrees. I don't whether this is right though.

Thanks CPhill i understand now!

Thank you so much!!!

Slope of line

[ 2 - -3 ]   / [ -5 -3 ]  =   5 / -8  =  -5/8

Equation of line  using (3 , -3)

y =(-5/8) ( x -3) - 3                multiply through by 8

8y  = -5 ( x -3) - 24

8y  = -5x + 15 - 24

5x + 8y  =  -9

A + B + C   =

5 + 8  + (-9)   =

4

Another  way

Construct  a circle at the origin with a radius of 1

Equation   is     x^2 + y^2  = 1   (1)

Slope  of line   y = x    =  1

So 

x = y

Sub this into  (1)  for y

x^2 + x^2  = 1

2x^2 = 1

x^2 = 1/2

x= 1/sqrt 2      and y =1/sqrt 2

Construct a perpendicular  line to y = x    passing through  (1/sqrt 2, 1/sqrt 2)

The equation is   

y = (-1) (x - 1/sqrt 2) + 1/sqrt2 = 

y =  -x + 2/sqrt2

y = -x + sqrt 2       (2)

Now consider the line y = (1/7)x

Sub this into (1) for y

x^2 + (1/7 x )^2   =1

x^2 + 1/49 x^2   = 1

(50 /49) x^2   =1

x^2  = 49/50

x = 7  /sqrt  (50)

y = (1/7) (7/sqrt (50)  = 1 /sqrt (50)

Now construct a perpendicular line  to y =(1/7)x    passing through  ( 7/sqrt 50 , 1/sqrt 50)

The equation  is

y = (-7)(x - 7/sqrt50) + 1/sqrt (50)

y = -7x + 50/sqrt50

y = -7x + sqrt (50)     ( 3)

Find  the  x intersection of   (2)  and (3)

-x + sqrt 2 =  -7x + sqrt 50

6x = sqrt (50)  - sqrt 2

6x  = 5sqrt (2) - sqrt 2

6x = 4sqrt 2

x = (2/3)sqrt 2

And  y =  - (2/3)sqrt 2  + sqrt 2  =   (1/3)sqrt 2

The intersection  of these two lines  is   [ (2/3)sqrt 2  ,  (1/3)sqrt 2) ] 

The slope of the line we  want (through BD)  =   [ (1/3)sqrt 2  ]  / [(2/3)sqrt 2 ]  =   1/2   =  .5 =  m

(Which we found by calculator  !!!)

please help somebody

Sorry......punched it into  the  calculator incorrectly.....see my corrected answer

Construct a circle  with  a center at the origin and a radius of 1

We have the equation 

x^2  +  y^2   =  1       (1)

Let P,Q lie on this  circle

OP has a slope of 2

So    y/x  =  2

y  = 2x

Sub this into (1)  for y

x^2 + (2x)^2   =1

5x^2  = 1

x^2  = 1/5

x =1/sqrt (5)

y = 2/sqrt (5)

So....P =  (1/sqrt (5)  / 2sqrt (5) )

OQ  has a slope of 3

So

y/x =  3

y = 3x

Sub this into (1) for y

x^2 + (3x)^2  =1

10x^2 = 1

x^2 = 1/10

x = 1/ sqrt (10)

y = 3/sqrt (10)

So....Q=  (1 / sqrt (10) , 3sqrt (10) )

Slope  of PQ  = 

[ 3/sqrt (10) - 2/sqrt (5) ]   /  [ 1/sqrt (10)   - 1/sqrt (5) ]   =

[ 3/sqrt (10)  - 2sqrt (2) / sqrt (10 ] /  [ 1/sqrt (10) - sqrt (2)/sqrt (10 ) ]  =

[ 3 - 2sqrt (2)]         [1 + sqrt (2)]          

____________ *    ____________    =

[ 1  -  sqrt (2) ]         [ 1 + sqrt (2)]

3 - 2sqrt (2) + 3sqrt (2) - 4

______________________   =

     1  - 2

-1 + sqrt (2)

__________  = 

    1  -  2

1  -  sqrt (2)  

Thanks, but that answer is incorrect. I haven't learned tan and arctan yet, do you have another solution? Thanks again.

Let  x = a + bi

Let y  = c -  bi

x + y = 2

a + c =  2

xy  =  (a + bi) (c  - bi)  = 

ac + (c - a)bi  + b^2  =  5

ac + 0bi + b^2  =  5

c- a  = 0

c = a  

a + c = 2

a + a = 2

2a = 2

a = 1  =  c

ac + b^2 = 5

1*1  + b^2  = 5

1 + b^2 =  5

b^2  = 4

b = 2     or  b  =  -2

( x + y)^2  =  2^2

x^2 + 2xy + y^2  = 4

x^2  + 2(5) + y^2  = 4

x^2 + y^2  =   -6

Let P=   (x,y)

PA^2  + PB^2 + PC^2  =

(x -2)^2 + (y + 5)^2  + (x -3)^2  + (y  -4)^2  + (x -1)^2 + (y + 2)^2  = 

3x^2 -12x + 3y^2 + 6y + 59  = 

3 ( x^2 - 4x + y^2  + 2y) + 59      complete the square on x ,y

3 [  x^2 - 4x + 4  + y^2 +2y + 1]  +  59  - 12 - 3

3 [ (x - 2)^2  + ( y + 1)^2 ] + 44

Q = (2, -1)

k =  44

To find the x coordinates of A, B

x^2 + (  [x - 4 ] / 2 )^2 =  6

x^2 + (1/4)(x^2 - 8x + 16)  =  6

4x^2 + x^2  - 8x + 16  =  24

5x^2 - 8x - 8   = 0

Sum of  roots =  8/5

x midpoint =   (8/5) /  2  =    8/10  = 4/5

y midpoint =  [ x - 4 ] / 2  =  [4/5 - 4 ] / 2  =  (-16/5) / 2  =  -8/5

Midpoint =  (4/5 , -8/5)

m  =  tan ( [ arctan (1)  + arctan (1/7) ]  /   2 )  =    .5

Corrected

i am finding answer of this question. 

TU * TW  = 10    →  TU =  10 / TW      (1)

TU * TP  =  16      (2)  

TP * TW =   30   →  TP  = 30 / TW     (3)

Sub (1) and (3) into (2)

10/TW  * 30 / TW  =  16

300 / TW^2  = 16

TW^2  =  300 / 16 =  75/4

TW = sqrt (25 * 3 ) / 2  = 5sqrt (3) / 2    = 2.5sqrt 3

TU  =  (10) /( 2.5sqrt 3)  =  4/sqrt 3

TP =  (30) / (2.5sqrt3)  =  12 / sqrt 3

Volume  = TU * TW *  TP  =   4/ (sqrt 3)  *  2.5(sqrt 3)  * 12 / (sqrt 3)  = 120sqrt 3 / 3 = 40sqrt 3

Corrected

sin a =     sqrt [ 5^2  - 3^2 ]  / 5 =   sqrt [ 16 ] / 5 =    4/5

cos B =   - sqrt [ 13^2 - 5^2 ] / 13 =  - sqrt [ 144 ] / 13  =   - 12/13

cos ( a - B)  = 

cos(a)cos(B) + sin (a) sin (B)  =  (3/5) (-12/13) + (4/5)(5/13)  = 

(-36 / 65)  + (20 / 65)    =

-16 / 65

No negative or 0  is in the  domain

Domain  = ( 0 , inf)

3/x  makes 0 not in the range

2 / (1 + 3/x)  =  2 / [ (x + 3) /x ]  =  [2x ] / (x + 3)    makes - 3 not in the range

1 + (2x) / ( x + 3)  =  [ (x + 3)+ 2x ] / (x +3) =  [ 3x + 3 ] / (x + 3)

So

1 / [ (3x + 3) / ( x + 3) ] =   (x + 3) / (3x + 3)      makes  -1  not in the  range

Sum of  three numbers =  0 + (- 3) + ( -1 )   =  -4