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Core Connections Algebra 2, 2013

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Core Connections Algebra 2, 2013 View details

1. Section 2.1

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Exercise 62 Page 78

Practice makes perfect

a We have a right triangle where both legs are known. Therefore, we can find the hypotenuse length, x, by using the Pythagorean Theorem.

a^2+b^2=c^2

SubstituteValues

Substitute values

6^2 + 5^2 = x^2

▼

Solve for x

Calculate power

36+25=x^2

Add terms

61=x^2

Rearrange equation

x^2=61

sqrt(LHS)=sqrt(RHS)

x=± sqrt(61)

x > 0

x=sqrt(61)

Notice that the side of a triangle is always positive which is why we disregarded the negative solution.

b Examining the diagram, we see that the hypotenuse and opposite side of ∠ C are known. With this information, we can set up a trigonometric equation containing ∠ C using the sine ratio.

By substituting the values for the opposite side and hypotenuse, we can determine the measure of C.

sin θ = Opposite/Hypotenuse

SubstituteValues

Substitute values

sin C=3/6

▼

Solve for B

a/b=.a /2./.b /2.

sin C=1/2

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

C=sin^(-1)1/2

300sin^(-1)(0)=0^(∘) 3030sin^(-1)(1/2)=30^(∘) 3045sin^(-1)(sqrt(2)/2)=45^(∘) 3060sin^(-1)(sqrt(3)/2)=60^(∘) 3090sin^(-1)(1)=90^(∘) 30-30sin^(-1)(- 1/2)=- 30^(∘) 30-45sin^(-1)(- sqrt(2)/2)=- 45^(∘) 30-60sin^(-1)(- sqrt(3)/2)=- 60^(∘) 30-90sin^(-1)(- 1)=- 90^(∘)

C=30^(∘)

c Like in Part B, we have a right triangle and want to find one of the angles. In this case we have bee given both of its legs which means we need to use the tangent ratio.

By substituting the values for the opposite side and hypotenuse, we can determine the measure of B.

tan θ = Opposite/Adjacent

SubstituteValues

Substitute values

tan B=4/5

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

B=tan^(-1)4/5

d In this triangle, we have been given the hypotenuse and one of the nonright angles. Since we want to find the measure of the angle's adjacent side, we can set up a trigonometric equation containing a using the cosine ratio.

By substituting the hypotenuse, adjacent side and the reference angle, we can determine the length of a.

cos θ =Opposite/Hypotenuse

SubstituteValues

Substitute values

cos 30^(∘) =a/10

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Solve for a

Rearrange equation

a/10=cos 30^(∘)

LHS * 10=RHS* 10

a= 10 cos(30^(∘))

\ifnumequal{30}{0}{\cos\left(0^\circ\right)=1}{}\ifnumequal{30}{30}{\cos\left(30^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{45}{\cos\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\cos\left(60^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{90}{\cos\left(90^\circ\right)=0}{}\ifnumequal{30}{120}{\cos\left(120^\circ\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{30}{135}{\cos\left(135^\circ\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\cos\left(150^\circ\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{180}{\cos\left(180^\circ\right)=\text{-} 1}{}\ifnumequal{30}{210}{\cos\left(210^\circ\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{30}{225}{\cos\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\cos\left(240^\circ\right)=\text{-} \dfrac {1}2}{}\ifnumequal{30}{270}{\cos\left(270^\circ\right)=0}{}\ifnumequal{30}{300}{\cos\left(300^\circ\right)=\dfrac{1}2}{}\ifnumequal{30}{315}{\cos\left(315^\circ\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\cos\left(330^\circ\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{30}{360}{\cos\left(360^\circ\right)=1}{}

a= 10 * sqrt(3)/2

MoveLeftFacToNum

a*b/c= a* b/c

a= 10sqrt(3)/2

Subchapter links

2.1.1 Modeling Non-Linear Data p.58-59

2.1.2 Parabola Investigation p.63-65

2.1.3 Graphing a Parabola Without a Table p.69-70

2.1.4 Rewriting in Graphing Form p.76-78

2.1.5 Mathematical Modeling with Parabolas p.81-82