Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 10.1
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Exercise 70 Page 521

Practice makes perfect
a

Examining the diagram, we can identify three types of wires. Each type has the same length. We have highlighted each type of wire in the diagram below based on their length.

Let's isolate two of the wires that connect to the midpoint of the roof's sides. Since they cut the length of each side in half, we can create two right triangles where both legs are known.

With this information, we can calculate the hypotenuse of each triangle by using the Pythagorean Theorem.

a^2+b^2=c^2
30^2+ 25^2=c^2
â–¼
Solve for c
900+625=c^2
1525=c^2
c^2=1525
c=± sqrt(1525)

c > 0

c=sqrt(1525)
c=sqrt(25* 61)
c=sqrt(25)sqrt(61)
c=5sqrt(61)

Using the Pythagorean Theorem again, we can calculate the length of the second wire.

a^2+b^2=c^2
16^2+ 25^2=c^2
â–¼
Solve for c
256+625=c^2
881=c^2
c^2=881
c=± sqrt(881)

c > 0

c=sqrt(881)

When we know the length of the wires that connect to the midpoint, we can use either of them to calculate the length of the wires that connect to the roof's four corners.

Let's use the Pythagorean theorem again.

a^2+b^2=c^2
30^2+( sqrt(881))^2=c^2
â–¼
Solve for c
900+881=c^2
1781=c^2
c^2=1781
c=± sqrt(1781)

c > 0

c=sqrt(1781)

Now we can calculate the total length of the wires by multiplying the length of each type of wire by how many there are in the picture and then adding the products.

2(5sqrt(61))+ 2(sqrt(881))+ 4(sqrt(1781))
306.27336...
306.27

b

From Part A, we know the length of the wire and of the antenna. Let's illustrate the angle we are looking for.

Since we know the hypotenuse and the opposite side to the angle with the roof, we must use the sine ratio to determine its measure.

sin θ =Opposite/Hypotenuse
sin θ =25/sqrt(1781)
â–¼
Solve for θ

sin^(-1)(LHS) = sin^(-1)(RHS)

θ =sin^(-1) 25/sqrt(1781)
θ =36.326825... ^(∘)
θ ≈ 36.32 ^(∘)

The angle is about 36.32^(∘)

c

If the antenna is x feet, we first have to redo all of the calculations from Part A that includes the height of the antenna.

30^2+x^2=( c_1)^2 ⇔ c_1=sqrt(900+x^2) 16^2+x^2=( c_2)^2 ⇔ c_2=sqrt(256+x^2)Next, we will calculate the length of the longest wire. 30^2+(sqrt(256+x^2))^2=( c_3)^2 ⇕ c_3=sqrt(1156+x^2) Now we can find the new length by adding c_1, c_2 and c_3. c_1+ c_2+ c_3 ⇕ sqrt(900+x^2)+ sqrt(256+x^2)+ sqrt(1156+x^2)