Exercise 34 Page 925

Function: d= 100/sin θ

Degree	Result
θ=60 ^(∘)	d ≈ 115.5
θ=50^(∘)	d ≈ 130.5

Practice makes perfect

We know that a radio tower has supporting cables attached to it at points 100 feet above the ground.

We want to write a model for the length d of each supporting cable as a function of the angle θ that it makes with the ground. As we can see, the tower and the cables form a right triangle with the ground.

Since we know the length of the height, we can use the trigonometric ratio for sine to find the expression for d. sin θ = Opposite/Hypotenuse In this case, the length of the opposite side is 100 and the length of the hypotenuse is d. sin θ = Opposite/Hypotenuse ⇒ sin θ = 100/d Let's solve this equation to obtain an expression for d.

sin θ = 100/d

MultEqn

LHS * d=RHS* d

dsin θ = 100

DivEqn

.LHS /sin θ.=.RHS /sin θ.

d= 100/sin θ

Now that we have the model for the length, we will substitute θ= 60^(∘) into this expression to find the value of the distance at this angle.

d= 100/sin θ

Substitute

θ= 60^(∘)

d = 100/sin 60^(∘)