The diagonals of a parallelogram are 12 meters and 20 meters and intersect at an angle of 60°. find the length of the longer side.

The diagram of the parallelogram is shown below

We want to work out the length of the longer side of the parallelogram. We can narrow down the shape into an obtuse triangle. We have the two sides of the triangle: 10 (from 20÷2) and 6 (from 12÷2) where 20 and 12 are the length of two diagonals. 

We will use the cosine rule c² = a² + b² - (2ab cos(C)) ⇒ refer to the second picture of the location of side a, b, c and angle C°

Angle C° is 120° (from the law of angles in a straight line add up to 180°)

So we have:
a = 10
b = 6
C° = 120°

Substitute into the cosine rule
c = 10² + 6² - (2×10×6×cos(120°))
c = 100 + 36 - (120 cos(120))
c = 136 - (-60)
c = 136+60
c = 196

c = x (refer back to the parallelogram diagram)

Hence, the longer side of the parallelogram is 196

The times are t = 9, t = 10, t = 11 and t = 12

Step-by-step explanation:

For the train to be more than 16 km South and since south is taken as negative,

d(t) > -16

t³ - 9t² + 6t > -16

t³ - 9t² + 6t + 16 > 0

Since -1 is a factor of 16, inserting t = -1 into the d(t), we have

d(-1) = (-1)³ - 9(-1)² + 6(-1)+ 16 = -1 - 9 - 6 + 16 = -16 + 16 = 0. By the factor theorem, t + 1 is a factor of d(t)

So, d(t)/(t + 1) = (t³ - 9t² + 6t + 16)/(t +1) = t² - 10t + 16

Factorizing  t² - 10t + 16, we have

t² - 2t - 8t + 16

= t(t - 2) - 8(t - 2)

= (t -2)(t - 8)

So t - 2 and t - 8 are factors of d(t)

So (t + 1)(t -2)(t - 8) > 0

when t < -1, example t = -2 ,(t + 1)(t -2)(t - 8) = (-2 + 1)(-2 -2)(-2 - 8) = (-1)(-4)(-10) = -40 < 0

when -1 < t < 2, example t = 0 ,(t + 1)(t -2)(t - 8) = (0 + 1)(0 -2)(0 - 8) = (1)(-2)(-8) = 16 > 0

when 2 < t < 8, example t = 3 ,(t + 1)(t -2)(t - 8) = (3 + 1)(3 -2)(3 - 8) = (4)(1)(-5) = -20 < 0

when t > 8, example t = 9,(t + 1)(t -2)(t - 8) = (9 + 1)(9 -2)(9 - 8) = (10)(7)(1) = 70 > 0

Since t cannot be negative, d(t) is positive in the interval 0 < t < 2 and t > 8

Since t ∈ (0, 12]

In the interval 0 < t < 2 the only value possible for t is t = 1

d(1) = t³ - 9t² + 6t = (1)³ - 9(1)² + 6(1) = 1 - 9 + 6 = -2

Since d(1) < -16 this is invalid

In the interval t > 8 , the only possible values of t are t = 9, t = 10.t = 11 and t = 12.

So,

d(9) = 9³ - 9(9)² + 6(9) = 0 + 54 = 54 km

d(10) = 10³ - 9(10)² + 6(10) = 1000 - 900 + 60 = 100 +60 = 160 km

d(11) = 11³ - 9(11)² + 6(11) = 1331 - 1089 + 66 = 242 + 66 = 308 km

d(12) = 12³ - 9(12)² + 6(12) = 1728 - 1296 + 72 = 432 + 72 = 504 km