Can you guys me find questions 6, 11, 13 & 19? ? i need to know the steps you took to solve it, or i'll fail geometry.

Question 6:
measure angle JKI + measure angle IKL = 180
Thus, measure angle IKL = 180 - 54 = 126 degrees
In triangle IKL, measure angle x + measure angle IKL = 54 degrees
since it is an isosceles triangle, thus:
measure angle x + measure angle IKL = 54 / 2 = 27 degrees
In triangle JIL:
measure angle y = 180 - (90 + 27) = 63 degrees

Question 11:
In the right triangle ABC, if = 90 and sin A = 5/13 , cos B is equal to ____________?
The sine of an angle = length of opposite side / length of hypotenuse
Therefore, we have a side length = 5 units and hypotenuse = 13 units
Use Pythagorean theorem to get the third side:
Third side = sqrt ( 13^2 - 5^2) = 12 units
The cosine of an angle = length of adjacent side /  length of hypotenuse
cos B = 4/13

Question 13:
angle A = 180 - (90 + 75) = 15 degrees
To get the lengths of the sides, we will use the sine law which can be written as shown in the attachment.
Thus,
(sin 90 / AB) = (sin 75 / 25) = (sin 15 / BC)
Doing the cross multiplication we get: BC = 6.698 units & AB = 25.882 units

Question 19:
In the 90-30-60 triangle, the length of the side opposite to 60 is [(root 3)/2] x the length of hypotenuse. 
cos 60 = 0.5 = length of adjacent / length of hypotenuse
hypotenuse = 8 units and the height of trapezoid = 6.928 units
length of ZX = 20 + 4 = 24 units
area of trapezpoid = [(20 + 24) / 2] x 4.6188 = 152.416 square units

Ithink so 1.you go to your profile and your questions and you'll be able to delete them