CPhill

Points (3,0)  and ( -1,-1) are on the  graph

Slope =  [ 0 - -1 ] / [ 3 - -1] = 1/4

Using (3,0)

y =(1/4) ( x -3)

y =(1/4)x - 3/4    multiply through by 4

4y = 1x -3

1x - 4y  = 3

Semi-perimeter =  [ 15 + 10 + 12 ]  / 2 =  18.5

Area  = sqrt  [18.5 * 3.5 * 6.5 * 8.5 ]  ≈  59.8

Area =  (1/2) (longest side )( altitude to this side)

 59.8  = (1/2) (15)  (altitude to this side)

59.8 / 7.5  ≈  7.97  =  shortest altitude

                           A

                 10               15

         B      1        D      3         C 

Let DC =  3BD

AD =  sqrt [ AC^2  - (3BD)^2 ]  = sqrt [ 15^2 - 9BD^2 ]      (1)

AD = sqrt [ AB^2 - BD^2 ]  =  sqrt [ 10^2 - BD^2 ]       (2)

Equate (1) , (2)

sqrt [ 15^2 - 9BD^2 ] = sqrt [ 10^2 -BD^2 ]              square both sides

15^2 - 9BD^2 = 10^2 - BD^2

225 - 9BD^2 = 100 - BD^2

125 = 8BD^2

125/8  = BD^2

AD = sqrt [ 10^2 - BD^2]  = sqrt [ 100 - 125/8 ] = sqrt [ 675/8]  = [15sqrt 3 ]  / [2 sqrt 2] =

7.5 sqrt [3/2]  =   7.5 sqrt (6) / 2   =  3.75sqrt (6) ≈ 9.19

                             A

                                     8

             B     2       K      3       C

Since AK is an  altitude

AK = sqrt [ AC^2 - KC^2 ]   =  sqrt [ 8^2  - 3^2 ] = sqrt [ 55]

And

AB = sqrt [ AK^2  + BK^2 ]  =  sqrt [ (sqrt 55)^2 + 2^2 ] = sqrt [ 59]

3.   Angle  B  = 180 - 55 -26  = 99°

10 / sin 99 = BC / sin  55

BC =  10sin55 / sin 99    ≈  8.3

2.

Call the point where the diagonals meet =  O

Triangle SOR  similar to triangle XOY

XY  / SR  = 5/28

So...height of triangle XOY  to height of triangle SOR = 5/28

Then there are 5 + 28  =  33  equal parts  in 1/2 height of the trapezoid

And triangle POQ similar to  triangle SOR

And height of POQ to height of SOR =    [ 5 +33] / 28  =  38/28   = 19/14

So

PQ  =  (19/14)SR=  (19/14) * 28 =   38 

1.

tan x + sec x =  2

sinx / cos x  + 1/cos x  = 2

sinx + 1   = 2cos x           square bith sides

sin^2 x + 2sin x + 1  = 4cos^2 x

sin^2 x + 2sin x + 1 =  4 (1-sin^2 x)

sin^2 x + 2sin x + 1 =  4 - 4sin^2 x

5sin^2 x + 2sin x - 3 =   0

(5sin x -3  )(sinx + 1)   = 0 

5sin x - 3  =  0                      sin x + 1  =  0

5sinx  = 3                              sin x  = -1

sin x =  3/5                                x = -pi/2     reject

cos x =   sqrt [ 5^2 - 3^2 ] / 5  =  sqrt [ 16 ] / 5  =   4 / 5

Very nice , Max !!!!

Let AO = x + r

OC = r

AC = 6

By the Pythagorean Theorem

r^2 + 6^2  = (r + x) ^2

r^2 + 36 = r^2 + 2rx + x^2

36 = 2rx + x^2

x^2 + 2rx  - 36   = 0

x^2 + 2rx  =  36     complete the square on  x

x^2 + 2rx + r^2  = 36 +r^2

(x + r)^2  = 36 + r^2

x + r = sqrt [36 + r^2 ]

x = sqrt [36 + r^2 ] - r

x  + 2r  =  12

sqrt [ 36 +  r^2 ] - r + 2r  = 12

sqrt [ 36 +r^2] = 12 - r

36 + r^2 = r^2 -24r + 144

24r = 108

r =108 / 24 =   9/2

Here :

https://web2.0calc.com/questions/function_66784