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

Exercise 122 Page 423

a Examining the diagram, we notice that we have a right triangle where we know a non-right angle and the hypotenuse. Since we want to find the length of the non-right angle's adjacent leg, we must use the cosine ratio.
Let's solve this equation.
cos 42^(∘) = x/10
Solve for x
10cos 42^(∘) = x
x=10cos 42^(∘)
x=7.43144...
x≈ 7.43
b From the diagram we see that the triangle is isosceles. Therefore, according to the Base Angles Theorem it has congruent base angles.
In any isosceles triangle the height, when drawn from the triangle's vertex angle, will bisect its base.
As we can see, x is now the hypotenuse of a right triangle. We also know a non-right angle and its adjacent leg. With this information we can write an equation containing x by using the cosine ratio. cos 45^(∘) = 7.5/x Let's solve this equation.
cos 45^(∘) = 7.5/x
Solve for x
xcos 45^(∘) = 7.5
x = 7.5/cos 45^(∘)
x = 8.061017...
x ≈ 8.06
c Like in Part A, we have a right triangle. In this case the hypotenuse and opposite leg to ∠ θ are known. With this information we can write an equation containing θ using the sine ratio.
To solve this equation, we have to take the inverse sine of both sides.
sin θ = 10/18
Solve for θ
sin θ = 5/9

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

θ =sin^(-1) 5/9
θ = 33.74898...^(∘)
θ ≈ 33.75^(∘)