CPhill

If I can read this...

angle B  =  62°

b  = 26.5

So A  =  90  - 62  =  28°

So......using the Law of Sines , we have

a/ sin A  =  b / sin B

a / sin 28 = 26.5 / sin 62        multiply both sides by sin 62

a  =  26.5 * sin 28  / sin 62  =   14.1 ft

The side opposite angle B  is  AC   =   "z"

The area, A, of a trapezoid is given by :

A =  height *  ( sum of the base lengths) / 2

So we have

123.265 = 8.9 * ( 17.8  + missing base)  / 2

Multiply both sides by 2

247.25  =  8.9 * (17. 8 + missing base ) 

Divide both sides by 8.9

27.78   =  17.8 + missing base   

Subtract 17.8  from both sides

≈ 9.98 miles  = missing base

The interest  is

Final value  - Amount invested  =

15000 (1.2)(6/3)  - 15000 =

15000 (1.2)(2) - 15000 =

15000 (2.4 - 1)  =

15000 (1.4)  =

$21000

Let's look at AZSun's answer a litlle more in detail.....

If Betty  and Wilma  join the group it is still not divisible into equal size groups

Note that all the even integers  from 30 - 50  can always be divided into equal groups because, at least, they are all divisible by 2

But look at the odd integers between  30 - 50

31   33    35    37   39    41   43   45   47  49

We are looking for two adjacent prime numbers

These  are   41  and 43

Note....41 students cannot be divided into equal groups and even if  Betty and Wilma join in....the group size  = 43    which isn't capable of being divided equally, either  !!!!

So...AZSun's answer of 41 is good  !!! 

1)

1- 3cosA - 4cos^2A                 1  - 4cosA

________________      =     ___________

       sin^2A                               1  - cos A

Note that we can factor the numerator on the left as

(1 - 4cos A) ( 1 + cos A)

___________________

  1  - cos^2A

(1 - 4cos A) (1 + cos A )

____________________

(1 - cos A) ( 1 + cos A)

1 - 4cosA

_________       which equals the right side

1  -  cos A

2)

tan A   =    1 + tan A

                 ________

                 1 +  cot A

One trick to try here, AT, is to write the right side in terms of sines and cosines

So we have

1 +  sin A /  cos A

_______________       get common denominators on top/bottom

1  + cos A / sin A

[cos A + sin A} / cos A

___________________      invert the bottom  fraction and multiply

[ sin A + cos A]  /  sin A

[ sin A + cosA ]           sin A

_____________   *    ____________

     cos A                   [  sin A + cos A]

sin A       *    [ sin A + cos A]

____             ____________

cos A          [ sin A + cos A ]

sin A       *  1

____

cos A

tan A  *  1   =   tan A     which is the left side     

Note...tertre....if we find the center of the circle and draw a radius  parallel to the bottom base of the large equilateral triangle, we will create another smaller equilateral triangle with a side of 6 cm

So.....the area of the sector formed by the two radial sides of this smaller equilateral triangle  will equal 1/6 of the circle whose radius  = 6  =

(1/6)pi (6^2)  =  6pi  cm^2        (1)

And the area of this smaller  equilateral triangle = 

(1/2)(6^2)*sqrt (3) / 2  =  

36sqrt (3) /4  =

9sqrt (3)  cm ^2       (2)

So....the   area of one of the shaded regions  is   (1)  - (2)  =

{ 6pi - 9sqrt (3)]  cm^2

So...by symmetry.....both shaded regions have an area of

2 [ 6pi - 9sqrt (3)]  cm^2  =

[ 12pi  - 18sqrt (3) ]    cm^2     

So

a + b + c  =

12 + 18 + 3

33

1) 

2sin^2A - 1  =   sin^4A - cos^4A

Working with the right side we can factor this as

(sin2^A + cos^2A) (sin^2A - cos^2A)  =

(1) ( sin^2 A - cos^2A)  =

sin^2A  -  (1  - sin^2A) =

sin^2A  - 1 + sin^2 A  =

2sin^2A  - 1        which is = to the left side

2) 

[ cos^2A - csc^2A + sin^2A]

     _______________________   =  - cot^3A

          sinAsec A

Obviously...the left side is the side that needs to be simplfied

Note......  sin^2A +cos^A  = 1

  [  1  - csc^2]

______________

   sinA (1/cos A]

Note

cot^2A  + 1  = csc^2A   ....so...

1 - csc^2A = - cot^2A

-cot^2A  * cosA

____________

   sinA

-cot^2A  *  cosA

                 ______

                  sin A

-cot^2A * cot A

-cot^3A       which is the right side

There are  46 total  students  in the class

Those that play football only  =  14 - 8  =  6

Those that play cricket only  = 20 - 8  =  12

So....those that play football,  cricket or both  = 6 + 12 + 8   =  26

So...the porbability that a student plays football, cricket or both is

26 / 46  =

13 / 23

We can find angle BCA  in the large triangle thusly :

sin BCA / 5.6  =  sin 50  / 9.3

sin BCA  =  5.6sin 50  / 9.3

So

arcsin (5.6 sin 50 / 9.3)  = BCA  =  27.47°

So....in the smaller triangle BDC, angle BDC  is supplemental to angle BDA  =  180 - 78  = 102°

So  x  = angle DBC  =

180 - BDC - BCA  =

180  - 102 - 27.47  = 

50.53° =  50.5°