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Common Core: High School - Geometry : Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Study concepts, example questions & explanations for Common Core: High School - Geometry

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Example Questions

Example Question #31 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Which of the following expressions properly states the Law of Sines? Use the following triangle for reference.

Screen shot 2020 08 13 at 8.30.42 am

Possible Answers:

$\frac{A}{cos(\gamma)} = \frac{B}{cos(\alpha)} = \frac{C}{cos(\beta)}$

$A^2 + B^2 - 2ABsin(\gamma) = C^2$

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$

$A^2 + B^2 -2ABcos(\gamma) = C^2$

Correct answer:

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$

Explanation:

The Law of Sines uses ratios of a triangle’s sides and angles to allow us to be able to solve for unknown sides and/or angles. This relationship is true for any triangle.

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Example Question #96 : Similarity, Right Triangles, & Trigonometry

Use the Law of Cosine to find the missing side. Round to the second decimal place.

Screen shot 2020 08 20 at 8.44.42 am

Possible Answers:

C = 3.57

C = 2.53

8.79

C = 5.40

Correct answer:

C = 5.40

Explanation:

We are able to use our equation from the Law of Cosine A^2 + B^2 - 2ABcos(c) = C^2 . We will start by assigning the values in our figure to the variables in the formula.

We will let:

A = 3

B = 4

c = 100

Now we can plug these values into our formula.

A^2 + B^2 - 2ABcos(c) = C^2

$\implies 3^2 + 4^2 - (2)(3)(4)cos(100) = C^2$

$\implies 9 + 16 - 24cos(100) = C^2$

$\implies 25 - (24)(-0.1736) = C^2$ $\implies 25 - (24)(-0.1736) = C^2$

$\implies 25 - (-4.1676) = C^2$

$\implies 29.1676 = C^2$

$\implies 5.40 = C$

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Example Question #97 : Similarity, Right Triangles, & Trigonometry

Using the Law of Sines, solve for and . Round to the second decimal place.

Screen shot 2020 08 20 at 10.01.08 am

Possible Answers:

x = 6, y = 6

x = 9.02, y =8.46

x = 6.72, y = 7.83

x = 10.29, y = 5.32

Correct answer:

x = 9.02, y =8.46

Explanation:

We begin by labeling our sides and angles according the variables in the equation for the Law of Sines: $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$

We will let

A = 6

B = x B = x

C = y

$\alpha = 75$

$\beta = 65$

$\gamma = 40$

We will begin by solving for . From our equation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ , we will leave out the term to solve for since it is not needed at this time.

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\implies \frac{6}{sin(40)} = \frac{x}{sin(75)}$

$\implies (6)sin(75) = (x)sin(40)$ (by cross-multiplication)

$\implies (6)(0.9659) = (x)(0.6428)$

$\implies 5.7954 = (x)(0.6428)$

$\implies 9.02 = x$

Now we will solve for . We could use $\frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ for our relationship, but we can also use $\frac{A}{sin(\gamma)} = \frac{C}{sin(\beta)}$ .

$\frac{A}{sin(\gamma)} = \frac{C}{sin(\beta)}$

$\implies \frac{6}{sin(40)} = \frac{y}{sin(65)}$

$\implies (6)sin(65) = (y)sin(40)$ (by cross-multiplication)

$\implies 5.4378 = (y)(0.6428)$

$\implies 8.46 = y$

So x = 9.02, y =8.46

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Example Question #93 : Similarity, Right Triangles, & Trigonometry

Use the Law of Sines to find the missing side lengths and angles. Round to the second decimal place.

Screen shot 2020 08 20 at 11.17.26 am

Possible Answers:

$C = 19.08, \alpha = 90, B = 10.15$

$C = 9.17, \alpha = 70, B = 15.63$

$C = 10.15, \alpha = 70, B = 19.08$

$C = 12.13, \alpha = 45, B = 23.45$

Correct answer:

$C = 10.15, \alpha = 70, B = 19.08$

Explanation:

We know for the Law of Sines our equation is $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac {C}{sin(\beta)}$ . We begin by assigning our values on the figure to variables in the equation.

We will let:

A = 20

$\gamma = 80$

$\beta = 30$

We will begin by solving for side .

$\frac{A}{sin(\gamma)} = \frac{C}{sin(\beta)}$

$\implies \frac{20}{sin(80)} = \frac{C}{sin(30)}$

$\implies (20)sin(30) = Csin(80)$

$\implies (20)(0.5) = C(0.9848)$

$\implies 10.15 = C$

Using the Law of Sines we need to solve for $\alpha$ and using the relation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$ . We do not know either or $\alpha$ . We are able to solve for $\alpha$ because we know that the angles of a triangle must add up to 180.

$30 + 80 + \alpha = 180$

$30 + 80 + \alpha = 180$ 70 = 180

Now we have enough information to solve for using the Law of Sines

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\implies \frac{20}{sin(80)} = \frac{B}{sin(70)}$

$\implies 20sin(70) = Bsin(80)$

$\implies (20)(0.9397) = B(0.9848)$

$\implies 18.7939 = B(0.9848)$

$\implies 19.08 = B$

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All Common Core: High School - Geometry Resources

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Common Core: High School - Geometry : Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #31 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #96 : Similarity, Right Triangles, & Trigonometry

Example Question #97 : Similarity, Right Triangles, & Trigonometry

Example Question #93 : Similarity, Right Triangles, & Trigonometry

All Common Core: High School - Geometry Resources

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