CPhill

         B

    6         8

A                   C 30

Law pf Cosines

AB^2  = AC^2 + BC^2  - 2(AC * BC) cos (30)

6^2 = AC^2 + 8^2  -  2 (AC * 8C) sqrt (3)/2

36  - 64 = AC^2   - 8sqrt (3) AC

-28 = AC^2 - 8sqrt (3) AC

AC^2  - 8sqrt 3 AC  + 28 =  0       solving for AC

AC   ≈ 2.46    or  AC ≈  11.4

So..because pf the SSA situation.....two possible areas result

[ ABC] =   (1/2) (2.46) (8) (1/2)  ≈ 4.92

[ABC ]  =  (1/2) (11.4) (8)(1/2)  ≈  22.8

The system is inconsistent

From the second equation , r =1

From the last equation r  = 9

Add the equations

2a = 10

a = 5

5 ( b + 1) = 250

b + 1 =  250 / 5

b + 1  =  50

b = 49

\(\frac{1}{64a^3 + 7} - 7 = 0 \)

1/[64a^3 + 7]   =  7

1 =  7 [ 64a^3 + 7]

1 = 448a^3 + 49

-48  = 448 a^3

[ -48 /448 ] = a^3

-3/28 = a^3

a =   - ∛ ( 3/28)

British Flag Theorem

AX^2 + CX^2  =  BX^2 + DX^2

3^2   + 3^2  = 3^2 + DX^2

3^2 = DX^2

3 = DX

ab^4 = 48           divide  the first wquation by the second

ab^2 = 4

b^2 = 12          take the both roots

b =  ± sqrt (12) = ± 2sqrt (3)

ab^2 =4

a (12) = 4

a = 4/12 = 1/3

This will be the difference in the volumes formed by 2 cones....see the foklowing

To find the radius of the cones, we need to find the   area of the triangle  (using Heron's formula)

Area =  sqrt  [ 14 *  6 * 6 *  2]  = 12sqrt 7

To find the  radius

12sqrt 7  =  (1/2) BC * height

12sqrt 7  = (1/2) (8) * height

3sqrt 7  = height  = radius  of both cones =  sqrt [ 63]  = AD

To find the height of one of the cones we have

sqrt [ 8^2 - 63 ] =  sqrt (1)  = 1

To find the height of the other cone we have 

sqrt [ 12^2 - 63 ] =sqrt (81)  = 9

Total volume  of rotated  triangle ABC =  Volume formed by rotation of triangle ADC - volume formed by rotation of triangle ABD  =

(1/3) pi (63) (9 - 1)  =  168 pi ≈ 527.79

Very nice......this one always stumped me  !!!!

Here's a similar problem :  https://web2.0calc.com/questions/help-pls_105

                   P

         5                   8

Q                   M                 R

            5.5                  5.5

Law of Cosines  (twice)

PR^2 =  PQ^2  + QR^2  - 2(PQ * QR)  cos (angle Q)

8^2  = 5^2 + 11^2 - 2(5 * 11) cos (angle Q)

[8^2 - 5^2  -11^2 ] / [ -110]  = cos (angle Q) = 41/55

PM^2  =  QM^2 + QP^2  - 2 ( QM * QP) cos (angle Q)

PM^2  =  (5.5)^2 + 5^2  - 2 ( 5.5 * 5) (41/55)

PM^2  = 14.25

PM  = sqrt (14.25) ≈  3.78