Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 12.2
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Exercise 61 Page 671

Practice makes perfect
a We want to determine the relationship between a and b from the following diagram.
Diagram of a circle

We see that a is the measure of an inscribed angle and b is the measure of a central angle. Both of these angles have the same intercepted arc.

Diagram of a circle

The measure of the central angle is the same as the measure of the intercepted arc. By the Inscribed Angle Theorem the measure of the inscribed angle is half the measure of its intercepted arc. This gives us the following relationship between a and b. a = 1/2 b Both a and b are positive, so this relationship tells us that a is less than b.

b We want to find the relationship between a and b from the following diagram.
Given triangle

The diagram shows a triangle with a and b as two of its sides. The angle opposite a has a measure of 62^(∘), but we do not know the measure of the angle opposite b. Once we know it, we can use the Law of Sines to relate a to b. So, first, let's find the measure of the last angle in our triangle. Let's call the measure of this angle x.

Given triangle with missing angle
In a triangle, the measures of a the internal angles sum to 180^(∘). We can write an equation for x using this fact. Let's do so and solve the resulting equation.
x + 62^(∘) + 70^(∘) = 180^(∘)
x + 132^(∘) = 180^(∘)
x = 48^(∘)
Let's add this measure to our diagram.
x = 48 degrees
Now, let's use the Law of Sines to relate a and b. sin 62^(∘)/a = sin 48^(∘)/b Now, let's solve this equation for a.
sin 62^(∘)/a = sin 48^(∘)/b
Solve for a
sin 62^(∘) = sin 48^(∘)/b a
sin 62^(∘)/.sin 48^(∘) /b. = a
sin 62^(∘) * b/sin 48^(∘) = a
sin 62^(∘)/sin 48^(∘) * b= a
a = sin 62^(∘)/sin 48^(∘) * b
a = 1.188123... b
We have that a = 1.188123... b. Since a and b are side lengths of a triangle, they are positive. Therefore, from a = 1.188123... b we conclude that a is greater than b.
c We want to find the relationship between a and b from the following diagram.
Given triangle
The diagram shows an equilateral triangle with a side length of 6 units and a square with a side length of 4 units. The area of the triangle is a, and the area of the square is b. We will find these areas, staring with the triangle.

Area of the Triangle

Let's recall the formula for an area A of an equilateral triangle with a side length s. A = sqrt(3)/4 s^2 Let's substitute 6 for s in that formula and simplify.
A = sqrt(3)/4 s^2
A = sqrt(3)/4 * 6^2
Simplify right-hand side
A = sqrt(3)/4 * 36
A = 36sqrt(3)/4
A = 9sqrt(3)/1
A = 9sqrt(3)
A = 15.588457...
A ≈ 15.6
The area of the triangle is equal to about 15.6 square units. We have that a is about 15.6. Now, let's move to finding the area of the square.

Area of the Square

A square with a side length s has an area s^2. In our case s= 4, so the area equals 4^2 = 16 square units. This means b = 16.

Comparison

We found that a ≈ 15.6 and b = 16. This means a is less than b.