CPhill

Since the diameter of the circle forms the largest chord possible in the circle.....we can position the triangle's  base along the diameter, D = 12 cm

The triangle with the greatest area will have the radius - R -  as its height

So.....the largest possible area of the prescribed triangle  =  (1/2) D * R    =  (1/2) (12) (6) =  36 cm^2

This was answered a while back, but the answer ends up being 16

4 ways to place the X in the top row

2 ways to place the X  in the second row [ in the other 2 x 2  box not occupied by the first X ]

2 ways to place the X in the  third row  [ we can choose either of the two empty columns]

Only 1 way  to place the last X in the last empty column

So   ....4 x 2 x 2 x 1   =  16

Again, because we have a bisected angle in a triangle....we have that

AX / XB =  AC /BC

AX / XB  =  28 / 56

AX/ XB  =  1 / 2      which implies that

AX : XB  =  1 : 2  

Then  AX + XB   make up three equal parts of AB

And AX  is 1/3  of the three equal parts....so...

(1/3) 50  = 50/3  = AX

Because angle BAC is bisected  by AD, we have that

BD / DC  =  AB / AC       so

42 / DC = 105 /AC     which implies that

105 * DC  =  42AC

DC  =  (42/105)AC =  (2/5) AC  =  .4AC

So....using the Pythagorean Theorem we have that

AB^2  + BC^2  = AC^2

AB^2  + ( BD + DC)^2  = AC^2

105^2  + ( 42 + .4AC) = AC^2   simplify

105^2  + 42^2 + 33.6AC + .16AC^2  =  AC^2  

12789 + 33.6AC + .16AC^2  =  AC^2    subtract the left side from both sides

.84AC^2   - 33.6AC - 12789 =  0

Using the quadratic formula will give us the positive solution for AC  =  145 

Here's my approach.....however.....there may be better methods of attack

Draw  altitude AD (through O ) of triangle ABC to base BC

This altitude will bisect BC  so that DC  = 3

The triangle ADC is right  and since AC = 5  and DC = 3, then this is a "Pythagorean" Triple" 3 - 4 - 5  right triangle......so.....AD  = 4

And  the triangle ODC formed is also right.... and  we have  that

 OD = √[OC^2 - DC^2]  =  √[OC^2 - 3^2] = √[ OC^2 - 9  ]  =  height of  ODC 

And since OC = OA....we have that

OA + OD  =   AD =  4      ....so...substituting

OC + √[ OC^2 - 9  ]  = 4     subtract OC from each side

√[ OC^2 - 9  ] = 4 - OC      square both sides

OC^2 - 9 =  16 - 8OC  + OC^2     subtract OC^2  from each side

- 9 = 16 -  8OC     subtract 16 from both sides

-25 =  -8 OC    divide both sides by -8

25/8 = OC

So the height of triangle OBC = OD  = √[ OC^2 - 9  ] = √[ (25/8)^2 - 9  ] 

And its area  =  (1/2) *BC * OD  =    

(1/2) * 6 * √[ (25/8)^2 - 9  ]   = 

3 * √ (625/ 64 - 576/64 )  =

3 * √ (49/64)  =

3 * 7/8  =

21 / 8  units^2

Since apex angle BCA is bisected we have the following relationship :

AX / BX  =  AC / BC

AX / 24  =  21 / 28

AX / 24  =  3 / 4      cross-multiply

4 *AX  = 24 * 3

4 * AX  =  72       divide both sides by 4

AX  =  18

Thanks, Melody.....other than exposing the forum members  [ and guests ] to, perhaps, a new concept, I can't really take much credit for the answers. I just ran across that website one day. Before that, I had never heard of this, either !!!!

[11x] / [3y^2] -  [ 7x] / [6y ]      the common denominator is 6y^2

[ 11x  * 2]  / [  3y^2 * 2 ]  -  [  7x * y ]/ [ 6y * y ]

[ 22x  -  7xy ] /  [ 6y^2 ]

-2*3.06+(7.6*6y^2)=240     simplify

-6.12 + 45.6 y^2  = 240     add 6.12 to both sides

45.6 y^2  =  246.12          divide both sides by  45.6

y^2  =  246.12 /  45.6  ≈   5.40

Note  that angle QRC  =  angle PRC  =  30°

So angle QRP  =  QRC + PRC  =  60°

So

QPR  =  [ 180  - QRP  -  PQR ] =  [ 180 -  60 - 65 ]  =  55°