A 20-meter long cable is used to support a telephone pole, holding the pole to make the pole perpendicular to the ground. If the cable forms a 60°
60
°
angle with the ground, how high up the pole should the cable be attached?

17.32m

Step-by-step explanation:

The shape formed by the arrangement between the pole, the ground and the wire is that of a right angled triangle.

Given that the cable forms 60° with the ground, the Pole is the opposite side of the triangle, the ground is the adjacent and the length of the supporting cable is the hypotenuse side.

Using the SOH CAH TOA notation, let the length of the point of attachment of the pole to the cable be x

Sin 60° = x/20

x = 20 Sin 60°

= 20 * 0.8660

= 17.32m

the answer to the expression is 83

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!