CPhill

We can find every angle in triangle PQR   [ using something known as the Law of Cosines ]

But....

Since we don't know what JL is......we cannot   find any of the  angles in triangle JKL

The center of the ellipse is ( 4, 1)

The major axis lies along x   and the minor axis lies along y

The equation of the ellipse  is

(x - 4)^2  / 16  + (y - 1)^2 / 1  = 1

c  =   sqrt (4^2 - 1)  =  sqrt (15)

And the distance between the foci = 2c  =  2sqrt (15)

tan (theta / 2)   =   tan (60/ 2)  =  tan (30)  =   1 /sqrt (3)  =  b / a

So

tan (60)  =  sqrt (3) / 1  =   a  / b

Note that y(3)  would mean that  we put 3 into y and evaluate

So......this would imply that

y(x)    means that we are putting  x into  y and evaluating.......so we have

5(3t^2 - t + 1)^2  - (5/3)(3t^2 - t + 1)    - 2  

I used Wolfram Alpha   to  get

45 t^4 - 30 t^3 + 30 t^2 - (25 t)/3 + 4/3

Angles R, S and T sum to 180°

Therefore

2x + 5 + 4x + 10 + 2x - 11 =   180  simplify

8x + 4  = 180       subtract 4 from both sides

8x  =  176        divide by 8

x =  22

Angle S  =  Angle X

4(22) + 10   =  98   =  Angle X

But A can't be correct because  9(22) / 2  = 99°

And C can't be correct because  (22)/2 + 22  =  33°

And  

Angle T  = Angle Y

2(22) - 11  =  33°

So

D  must be correct because   (22)/2 + 22  =  33°

So  

Angle T = Angle Y

If a^2 + b^2  >  c^2

The triangle is acute

If   a^2 + b^2  = c^2

The triangle is right

If a^2 + b^2 < c^2

The triangle is obtuse

D)One pair of corresponding sides has the same measure.

This would allow us to use   ASA   congruency

2a

He would need to know that angle JKL  = angle PQR

2b

He would need to know that JL  = 20

I'll do the first one.....see if you can do the second

1.   

First term  = 2(3) - 10  =  - 4

Last term = 2(6) - 10 = 12

Sum   = ( first term + last term) * (number of terms)  / 2   = (-4 + 12) (6 - 3 + 1) /2  =

(-4 + 12) (4) /2  =

8 ( 4) / 2 =

32 / 2  =

16

Second one

First term =  2(1)^2 - 4  =  -2

Last term = 2(4)^2 - 4  =  32 - 4  =  28

Number of terms  =  4

The second answer