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H will be the orthocenter of  the triangle   

Let  CZ  be the other altitude

Then we have right triangle  CZB  with  angle CZB  = 90  angle ZBC  = 82   and angle ZCB   = 180 - (90 +82)  =

8°

And we have another right triangle CHX  with angle HXC = 90  and angle HCX  = 8

So angle CHX  =  180 - (90 + 8)   =  180   - 98   =  82°

thank you so much melody!!!

Please proofread your qwuestions before you post them

We can find the height of the  cylinder as follows

60 =   pi * 2^2  * height

15 / pi  =  height

Now.....the radius  of  the sphere  can  be found by the Pythagorean Theorem

We have a right triangle with one leg  =  the radius of the cylinder = 2  and the other  =  (1/2) the height of the cylinder =  7.5 / pi

The radius of the sphere is  the hypotenuse of this  triangle   = sqrt  [ 2^2 + (7.5/pi)^2  ]  = 

sqrt  [ 4 + 56.25/pi^2 ]  =   [ 4 + 56.25 / pi^2]^(1/2)  

The   volume of this  sphere  =    (4/3) pi * [  4 + 56.25/pi^2] ^(3/2)  ≈  126.5  cm^3

Because of  symmetry, the area of triangle ACF will be the same as the area of triangle EAC

EC  is the base  =  the diagonal of a square with a side of 3  =    3sqrt 2

The height of this triangle  =    sqrt   [3^2   - [(3/2)sqrt2]^2 ]  =   sqrt  [  9   - 9/2 ] = sqrt [ 9/2]  = 3 /sqrt 2

So.....the area of triangle  ACF =   (1/2) [3sqrt 2] [3 /sqrt 2 ] =   9/2  = 4.5 units^2  

choose a point, from there 9 lines can be draw

go clockwise to the next point. 8 new lines will be able to be drawn

etc until only one new line can be drawn

so that would be    9+8+7+6+5+4+3+2+1

That is wrong, sorry, I don't know how to start

the answer is 0 Cdortch

i am

t E (pretend its the curvy E) (-9/8, 3/2)

\(\mathrm{Use\:the\:following\:identity}:\quad \sin \left(s\right)-\sin \left(t\right)=2\cos \left(\frac{s+t}{2}\right)\sin \left(\frac{s-t}{2}\right)\)

\(\sin \left(88^{\circ \:}\right)-\sin \left(32^{\circ \:}\right)=2\cos \left(\frac{88^{\circ \:}+32^{\circ \:}}{2}\right)\sin \left(\frac{88^{\circ \:}-32^{\circ \:}}{2}\right)\)

\(2\cos \left(60^{\circ \:}\right)\sin \left(28^{\circ \:}\right)\)

\(2\cdot \frac{1}{2}\sin \left(28^{\circ \:}\right)\)

\(\sin(28)\)

-Vinculum

If the  perimeter of the square =  60, then each side = 15

The area of the square  =  15^2  = 225

And we have 4 congruent  right triangles with legs = 5  formed  outside the octagon but inside the square

The  area  of each = (1/2)(5)^2  =  25 / 2

So  their  combined area  = 4 * 25  / 2  =  50

So  the area of the octagon =   225  -  50  =    175 

Let the side of  the square  =   x

Area of triangle   BCE  =    (1/2)   (x) (CE)

                      3   =    (1/2) (x) (CE)

                      6 / x  = CE

                    So  ED  = ( x - CE ) =  (  x - 6/x) 

Area of triangle   BAF  = (1/2) (x) (AF)

                         4 =  (1/2) (x) (AF)

                           8/x = AF

                      So  FD  =  (x - AF)  = ( x - 8/x) 

And area of  FED  =  (1/2)(FD)(ED)

         5 =  (1/2) (x - 8/x) (x - 6/x)

        10 =   (x^2 - 8) (x^2 - 6) / x^2

         10x^2  = x^4 - 14x^2  +  48

So

x^4  - 24x^2  +  48   = 0              complete the square

x^4  - 24x^2  + 144  =    -48 + 144

(x^2  - 12)^2  =  96

x^2  - 12  = sqrt (96)

x^2  =  sqrt (96)  + 12  =    4sqrt 6 + 12  = 4 ( sqrt 6 + 3)  =   area of ABCD  ≈ 21.8