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Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Your friend wants to build a triangular mirror with a wooden frame for her living room. She already has two cut wooden pieces that are 6 inches in length. She has put these sides together at an angle of 45 degrees. What length should she cut the third side to be? Round to the second decimal place.

1.32 $\displaystyle 1.32$

3.78 $\displaystyle 3.78$

4.59 $\displaystyle 4.59$

7.95 $\displaystyle 7.95$

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Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

1.32 $\displaystyle 1.32$

3.78 $\displaystyle 3.78$

4.59 $\displaystyle 4.59$

7.95 $\displaystyle 7.95$

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Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Your answer:

1.32 $\displaystyle 1.32$

Your answer:

3.78 $\displaystyle 3.78$

Your answer:

7.95 $\displaystyle 7.95$

Correct answer:

4.59 $\displaystyle 4.59$

Explanation:

Since we know two sides and their adjoining angle, we are able to use the Law of Cosine to solve for the third side. Recall that the formula for the Law of Cosine is $\displaystyle A^2 + B^2 - 2ABcos(c) = C^2$ . We will begin by defining our variables according to the problem.

We will let

A = 6 $\displaystyle A = 6$

B = 6 $\displaystyle B = 6$

c = 45 $\displaystyle c = 45$

How do we know that $\displaystyle c$ is the angle we should use? $\displaystyle c$ is the angle that would be directly opposite of side $\displaystyle C$ (our unknown side) so this is our desired angle to use in the formula. Now we will plug these values into our formula.

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

$\implies 6^2 + 6^2 - (2)(6)(6)cos(45) = C^2$

$\implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$

$\implies 72 - 36\sqrt{2} = C^2$

$\implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$ $\implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$ $\implies 21.0883 = C^2$ $\displaystyle \implies 21.0883 = C^2$

$\implies 4.59 = C$ $\displaystyle \implies 4.59 = C$

Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

You are given a triangle with 2 known sides and 2 known angles. Which method could you use to solve for the unknown side and angle.

Both the Law of Cosines and Law of Sines

Just the Law of Sines

Neither the Law of Cosines and Law of Sines

Just the Law of Cosines

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Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

You are given a triangle with 2 known sides and 2 known angles. Which method could you use to solve for the unknown side and angle.

Your answer:

Just the Law of Sines

Your answer:

Neither the Law of Cosines and Law of Sines

Your answer:

Just the Law of Cosines

Correct answer:

Both the Law of Cosines and Law of Sines

Explanation:

Take the triangle below for example:

First, we can figure out the third angle simply because we know all three angles of a triangle must add up to be 180. So

$60 + 60 + \gamma = 180$ $\displaystyle 60 + 60 + \gamma = 180$

$\implies \gamma = 60$ $\displaystyle \implies \gamma = 60$

Now we have enough information to use either law. We will begin by using the law of cosine $\displaystyle A^2 + B^2 - 2ABcos(c) = C^2$ .

We will let

A=7 $\displaystyle A=7$

B=7 $\displaystyle B=7$

C=60 $\displaystyle C=60$

Now we will plug these values into our formula

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

$\implies 7^2 + 7^2 - (2)(7)(7)cos(60) = C^2$

$\implies 49 + 49 - 98(\frac{1}{2})= C^2$

$\implies 98 - 49 = C^2$ $\displaystyle \implies 98 - 49 = C^2$

$\implies 49 = C^2$ $\displaystyle \implies 49 = C^2$

$\implies 7 = C^2$ $\displaystyle \implies 7 = C^2$

We can also prove that using the Law of Sines will give us the same answer. Our formula for the Law of Sines is $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ $\displaystyle \frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ .

We will let:

A=7 $\displaystyle A=7$

C=7 $\displaystyle C=7$

$\alpha = 60$ $\displaystyle \alpha = 60$

$\beta = 60$ $\displaystyle \beta = 60$

$\gamma = 60$ $\displaystyle \gamma = 60$

We can set up with the relation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$ $\displaystyle \frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

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

$\implies \frac{7}{sin(60)} = \frac{B}{sin(60)}$

$\implies 7sin(60) = Bsin(60)$ (by cross-multiplication)

$\implies \frac{7sin(60)}{sin(60)} = B$

$\implies 7 = B$ $\displaystyle \implies 7 = B$ ( sin(60) $\displaystyle sin(60)$ cancels)

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Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

All Common Core: High School - Geometry Resources

Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Flashcards: Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

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