Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
Chapter Closure

Exercise 116 Page 477

a Two of the triangle's angles are 60^(∘). By the Triangle Angle Sum Theorem, we can determine the third angle.

θ + 60^(∘)+60^(∘)=180^(∘) ⇔ θ=60^(∘) Since all of the triangle's angles are 60^(∘), this is an equilateral triangle. In an equilateral triangle, all three sides have the same length. Therefore, we know that x=8 cm.

b From the diagram, we have been given the adjacent leg to the reference angle and are asked to find the length of the hypotenuse. With this information, we can use the cosine ratio to determine x.
cos θ = Adjacent/Hypotenuse

Let's look at the given diagram and substitute the given values into the above equation.

Let's solve this equation.
cos(29 ^(∘)) = 27/x
Solve for x
x cos(29^(∘)) = 27
x = 27/cos(29^(∘))
x=30.8705598326...
x≈30.87
c Let's take a look at the given diagram.

It depicts a right triangle in which the hypotenuse and one of the legs is given. We want to find the length of the other leg, which we can do using the Pythagorean Theorem.

Pythagorean Theorem

In a right triangle, the length of the hypotenuse squared equals the sum of the squares of the lengths of the legs. a^2+b^2=c^2

Let's substitute the given values into the above equation and solve for x.
a^2+b^2=c^2
5^2+ x^2= 16^2
Solve for x
25+x^2=256
x^2=231
x=± sqrt(231)

x > 0

x=sqrt(231)
x=15.19868...
x≈ 15.20
d From the diagram, we have been given the opposite leg to the reference angle and are asked to find the length of the adjacent one. With this information, we can use the tangent ratio to determine x.
tan θ = Opposite/Adjacent

Let's look at the given diagram and substitute the given values into the above equation.

Let's solve this equation.
tan(58^(∘))=6/x
Solve for x
x tan(58^(∘)) = 6
x = 6/tan(58^(∘))
x=3.74921611146...
x≈3.75
e Let's take a look at the given diagram.

It depicts a right triangle in which the the two legs have the same length of 7 units. We want to find the length of the hypotenuse, which we can do using the Pythagorean Theorem.

Pythagorean Theorem

In a right triangle, the length of the hypotenuse squared equals the sum of the squares of the lengths of the legs. a^2+b^2=c^2

Let's substitute the given lengths into the above equation and solve for x.
a^2+b^2=c^2
7^2+ 7^2= x^2
Solve for x
49+49=x^2
2(49)=x^2
sqrt(2(49))=x
sqrt(2)sqrt(49)=x
sqrt(2)(7)=x
7sqrt(2)=x
x = 7sqrt(2)