The width of a rectangle is 5 feet more than 3 times the length. the perimeter is 314 feet. find the length and the width.

We know that the perimeter of a rectangle is two times the width plus two times de length, therefore the result is:

Length = 38 feet

Width = 119 feet

Step-by-step explanation:  

Being width  w , the length L   and the perimeter P , we have then:

w = 5 feet + 3L  (1)

For the perimeter:

P = 2w +2 L    (2)

Replacing (2) on (1)

P = 2(5 feet + 3L) +2 L

As we know the perimeter P =314 feet, thus:  

P = 2 (5 feet + 3L)+2 L

P = 10 feet + 8 L = =314 feet

Clearing L,  we have the length is:

L = 38  feet (3)

Replacing (3)  on (1), we have the width is:

w  = 119 feet

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)