Two guy wires from the top of a 200-foot radio antenna are anchored to the ground by two large concrete posts. The antenna is perpendicular to the ground and between the two posts. If the angles of depression from the top of the antenna to each of the posts is 46o and 48o respectively, then the distance between the two posts is:

373.22 feet

Step-by-step explanation:

The attached photo contains the diagram illustrating the scenario. 2 right angle triangles ACD and BCD are formed.

The distance between the two posts is AD + BD

To determine AD, we would apply the tangent trigonometric ratio which is expressed as

Tan θ = opposite side/adjacent side

Tan 46 = 200/AD

AD = 200/Tan 46 = 200/1.0355

AD = 193.14

To determine BD, we would also apply the tangent trigonometric ratio. Therefore

Tan 48 = 200/BD

BD = 200/Tan 48 = 200/1.1106

BD = 180.08

Therefore, the distance between the two posts is

193.14 + 180.08 = 373.22 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)