Core Connections Integrated II, 2015
CC
Core Connections Integrated II, 2015 View details
1. Section 10.1
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Exercise 46 Page 562

Practice makes perfect
a

To calculate the measure of an interior angle of a regular n-gon we can use the formula 180^(∘)(n-2)n, where n is the number of sides in the polygon. A pentagon is a polygon with 5 sides.

180^(∘)(n-2)/n
180^(∘)( 5-2)/5
â–¼
Evaluate
180^(∘)(3)/5
540^(∘)/5
108^(∘)

The interior angle of the regular pentagon is 108^(∘).

Next, we will draw an equilateral triangle with a central angle of a.

The sum of the central angles in any regular polygon equals 360^(∘). Since an equilateral triangle has 3 congruent central angles, we can determine the measure of a by dividing 360^(∘) by 3. 360^(∘)/3= 120^(∘) As we can see, a=120^(∘) and b=108^(∘). Therefore a is greater than b.

b

Notice that a and b are corresponding angles. If the two lines cut by the transversal were parallel, we could claim that a=b by the Corresponding Angles Theorem.

However, we do not have this information. Therefore, we have no way of saying which angle is greater or if they are the same.

c

With the given information, we can calculate ∠ a with the sine ratio and ∠ b with the cosine ratio.

Let's solve these equations for a and b.

sin a=8/12
â–¼
Solve for a
sin a=2/3

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

a=sin^(-1)2/3
a=41.8103148
a≈ 41.81^(∘)

Let's also solve for b in the second equation.

cos b=8/12
â–¼
Solve for b
cos b=2/3

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

b=cos^(-1)2/3
b=48.18968...
b≈ 48.19^(∘)

From our calculations, we see that b is greater than a.

d

The equation tells us that to obtain a we have to add 3 to b. This must mean that a is greater than b.

a = b + 3 a is 3 greater than b
e

Like in Part A, we can use trigonometric ratios to write equations containing a and b. For the triangle on the left we need to use the cosine, and in the triangle on the right we need to use the sine ratio.

Notice that both triangles have the same hypotenuse. It will be easier to compare a and b if we express their values using the same trigonometric function. To do so, let's find the measure of the missing angle in the right triangle. Ratio b 7 could be then expressed using the sine ratio.

In a right triangle two acute angles are complementary. Therefore, the sum of their measures will add up to 90^(∘). This allows us to write an equation that we can solve for x. x + 67^(∘) = 90^(∘) ⇔ x = 23^(∘) The measure x is the same as the measure of the acute angle in the left triangle. Therefore, we can equate the sine ratios. a/7 = sin 23^(∘) = b/7 We can multiply the equation between the outward terms by 7. a/7 = b/7 ⇔ a = b We see that a is equal to b.