geno3141

C = 85.00 + 0.48·d

Draw altitude AP with P on BC.

Since B = arcsin(24/25), sides AP and AB ( of triangle(APB) ) must be in the ratio 24/25.

Since C = arcsin(12/13), it is also true that C = arcsin(24/26) and that sides AP and AC

( of triangle(APB) ) must be in the ratio 24/26.

In triangle(APB), if AP = 24 and AB = 25, then BP = 7 (Pythagorean Theorem).

In triangle(APC), if AP = 24 and AC = 26, then CP = 10.

Since BP + CP = BC and since 7 + 10 = 17, BP must equal 10 and CP must equal 7.

The area of triangle(ABC) = ½·base·height =  ½·BC·PA = ½·17·24 = .........

angle(A) = 46.8.

sin( angle(A) ) = a / c   --->   sin( 46.8 ) = 3.2 / c   --->   c  =  3.2 / sin( 46.8 )  =  4.4

AB = 4.4.

AC can't be larger than AB, so ...

The two triangles, EBQ and CDQ are similar by A-A-A.

Thus, the areas of the triangles are in the ratio of the squares of the corresponding sides.

Since 3BE = 2DC, BE = (2/3)DC and the ratio of the areas of triangle(EBQ) and triangle(CDQ)

is 4/9.

Since triangle(QDC) has area 36, triangle(EBQ) has area (4/9)·36  =  16.

Assuming that P, Q, and R are on BC (in the order B, P, Q, R, C):

Angle(BAC) = 90.

Since AQ is an angle bisector, angle(BAQ) = 45.

Since P is between B and Q and angle(PAQ) = 13, angle(BAP) = angle(BAQ) - angle(PAQ)

              --->   angle(BAP) = 45 - 13 = 32.

Since AP is an altitude, angle(BPA) = 90.

In triangle(BAP), angle(BAP) = 32 and angle(BPA) = 90   --->   angle(ABP) = 58.

In triangle(BAC), angle(BAC) = 90 and angle(ABP) = 58   --->   angle(ACB) = 32.

Since AR is a median of a right triangle, RA = RB = RC.

Since RA = RC, triangle(ARC) is an isosceles triangle.

Since angle(ACB) = 32, ange(CAR) = 32.

The figure on the left:  

Because the lines are parallel,

   (5y - 4) + (3y)  =  180^o   --->   8y - 4  =  180^o   --->   8y  =  184^o   --->   y  =  23^o  

   3y  =  2x + 13   --->   3(23)  =  2x + 13   --->   69  =  2x + 13   --->   56  =  2x    --->   28^o  =  x

The figure on the right:

Because the lines are parallel,

   3x + 2y + 4y  =  180^o   --->   3x + 6y  =  180^o  

Because they are in a triangle,

   2y + 4y + 5x - 20^o  =  180^o   --->   6y + 5x  =  200^o 

Solving these equations:  6y + 3x  =  180^o 

                                         6y + 5x  =  200^o

Subtracting:                            -2x  =  -20^o

And, now it's your turn ...

Knowing that  (n + 1)!  =  (n + 1)·n!:

2(n + 1)!  +  6n!  =  3(n + 1)!

2·(n + 1)·n!  +  6·n!  =  3·(n + 1)·n!

Divide both sides by  n!:

2(n + 1)  +  6  =  3(n + 1)

    2n + 2 + 6  =  3n + 3

          2n + 8  =  3n + 3

                  8  =  n + 3

                  5  =  n

Use the identity:  cos²(x)  =  1 - sin²(x)  so that the equation only conains the  sin  function.

cos²(x) + 3sin(x) - 3  =  0     --->     ( 1 - sin²(x) ) + 3sin(x) - 3  =  0

                                                                -sin²(x) + 3sin(x) - 2  =  0

                                                                  sin²(x) - 3sin(x) + 2  =  0

                                                            ( sin(x) - 2 )( sin(x) - 1 )  =  0

                     either     sin(x) = 2  [impossible]     or     sin(x) = 1   --->   x = pi/2

One way is to list all the possibilities for division by 4, for division by 5, and for division by 6, until you get the same number in each list.

The integers that, divided by 4, leave a remainder of 3 are:  7, 11, 15, 19, ...   (notice that these numbers increase by 4)

The integers that, divided by 5, leave a remainder of 3 are:  8, 13, 18, 23, ...   (these numbers increase by 5)

The integers that, divided by 6, leave a remainder of 3 are:  9, 15, 21, 27, ...    (these numbers increase by 6)

Extend each list until you get the same number in each list (the answer is smaller than 125).

There is a theorem in geometry that states that:  (8)(8 + 2)  =  (x)(x)

If you use the other line segment:  (4)(4 + 16)  =  (x)(x)

Either way, you end with:  80  =  x²   --->   x  =  sqrt(80)