CPhill

First one  ....Note that

sin L  = 21/75  ⇒    sin^-1 (21/75)   = L

cos L  = 72/75 ⇒  cos-1 (72/75)  = L

tan L = 21/72  ⇒  tan-1 (21/72) = L

Second one

sin A  = 35/37

cos A  = 12/37

Task 3

y  =   (x + 3)

        ______

         (x - 2) 

The horizontal asymptote  occurs at  y  = 1

The vertcal asymptote occurs at  x  = 2

The x intercept is  at  x  = -3

The y intercept is  at   (-3/2)

Here's a  graph : 

https://www.desmos.com/calculator/4l3egciy58

Here's a few

Task 2

y  = x^2       (quadratic)

y = x            (linear)

Sub  the   second equation into the first for y

x  = x^2       subtract x from  both sides

0  = x^2  - x      

x^2  - x   =  0   factor

x ( x - 1)  =0

Set each factor to 0  and solve for  x

x  = 0        x  - 1  =  0   

                add 1 to  both sides

x = 0          x  = 1

So....when x  = 0, y  = 0

And when x  = 1, y  = 1

So....the solutions are  (0,0)  and (1,1)

Here's a graph ;

https://www.desmos.com/calculator/ldpimov7kd

First one  ...this is a  30-60-90 right triangle

The side opposite  the  60° angle will be  √3 * times the side opposite the 30° angle  = √3 * 8  = 8√3  =  √192

So....you were correct....!!!!!!!

Second one

We can use the tangent, here ;

tan (20°) = 4 / x         we can rearrange this as

x  =  4  / tan(20°)  = 10.99

Let  O  = (0,0)

Let A  = (2,3)

Let B = (1,1)

Using the "tip to tail" method....let   C  = (2 + 1, 3 + 1)  = (3,4)

So......the length of diagonal  OC  =   √ [ (3 - 0)^2  + (4-0)^2  ] =   √[9 + 16 ] = √25  =  5   units

And the length of diagonal AB  = √ [ (2-1)^2 +  ( 3 - 1)^2 ]  = √ [ 1^2  + 2^2]  =  √5 units

The perimeter  is   2*OA  = 2 * √ [ 2^2 + 3^2 ] = 2 *√13    +    2* AC  = 2* √ [ (2 - 3)^2 + (3 -4)^2]  = 2 √2

So....the total perimeter is     2  [√13  + √2 ]  ≈  10.04  units

Here's a pic :

Remember that    a(x1, y1)   - b(x2, y2 ) =   ( ax1 - bx2, ay1  - by2 )

a(3,4)  - b(2,4)   =  (8,10)

So we have that

3a  - 2b  = 8      and

4a  - 4b  = 10

Multiply the first equation by - 2  all the way through

-6a  + 4b  =  -16     

4a  -   4b   =  10        add these

-2a   =  -6     divide both sides by  -2

a  = 3

And using   3a  - 2b  = 8  to  find b, we have that

3(3)  - 2b  = 8

9  - 2b  = 8 

-2b  = -1

b  = 1/2

OK....no prob   !!!!

Here's one possibility  [ but maybe not the only one ]

The sum of the interior angles  = 540°

Let the smallest angle  = 60°

Let the next two angles   =  60.5°    and 61.5°

Note that the median  =  [ 60.5  + 61.5)  / 2  = 61°

Thw sum of the angles, so far, is  182° 

So...the sum of the other two angles will be     540  - 182  =  358°

Let the next two angles  = 179°   

The sum of the angles is 

[60 + 60.5 + 61.5 + 179 + 179]  =  540°

The median  =  61°    and   there is only one mode,  179°

First one :

We know an opposite side and an adjacent side....the tangent seems most appropriate, here

So we have

tan x  = 11/9       use the tan  inverse  (atan) to find x

atan (11/9)  =  x ≈  50.71° = 51°

Second one....

tan Y  = 6/7          tan Z  = 7/6

Third  one :

We know an angle, the hypotenuse and we are looking for an adjacent side to the angle....the cosine is what we need

cos (26)  = x / 10   multiply both sides by 10

10 cos (26)  = x   ≈  9

Fourth one :

We know an angle, the adjacent side....and we are looking for the hypotenuse....the cosine, again....

cos36  = 12 / x       divide both sides by cos 36......multiply both sides by  x

x   =  12  / cos 36  ≈  14.8

Last one

We know an angle, an adjacent side....and we are looking for the opposite side.....the tangent seems good

tan  40  = x / 5       multiply both sides by  5

5 tan (40)  = x  ≈  4.2

y = ax^2 + b        y  = 2x^2 + cx

Since the point (-1, 0)   is on both graphs

Using the first function, we have

0 = a(-1)^2 + b

0  =  a  + b

Using the second function, we have that

0 = 2(-1)^2 + c(-1)

0 = 2  - c

c = 2

Taking the derivative of both functions

y'  = 2ax               y'  = 4x + c

y' = 2ax               y' =  4x + 2

The slope of the tangent line is the same for both functions at x  = -1

So....equating slopes

2a(-1)   =  4(-1) + 2

-2a  =    -4 + 2

-2a  = = -2

a  =  1

And since a + b  = 0

b  =  -1

The equation of the tangent line is

y =  (4(-1) + 2) ( x  - -1)

y =  -2(x + 1)

y =  -2x -2

Here's a graph  :  https://www.desmos.com/calculator/2wiesd0ekz