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# Common Core Standard HSG-SRT.D.10 Questions

(+) Prove the Laws of Sines and Cosines and use them to solve problems.

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Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing reason in step 13?
1. Distributive Property
2. Substitution Property of Equality
3. Pythagorean Identity
4. Pythagorean Theorem
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing statement in step 5?
1. $sin(m ang B) = (AD)/(AB)$
2. $sin(m ang B) = (AB)/(AD)$
3. $sin(m ang B) = (BD)/(AB)$
4. $sin(m ang B) = (BD)/(AD)$
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
In $Delta XYZ$, $XY = 12 \ "units"$, $YZ = 15 \ "units"$, and $m ang X = 50°$. Using the law of sines and the given information, which of the following is true?
1. No such triangle exists.
2. These measurements result in a unique triangle.
3. These measurements result in two possible triangles.
4. The law of sines cannot be used in this situation.
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing reason in step 18?
2. Pythagorean Identity
3. Pythagorean Theorem
4. Algebra (collect like terms)
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing statement in step 7?
1. $sin(m ang C) = (AB)/(BC)$
2. $sin(m ang C) = (CD)/(AC)$
3. $sin(m ang C) = (AD)/(CD)$
4. $sin(m ang C) = (AD)/(AC)$
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
In $Delta QRS$, $QR = 9 \ "units"$, $RS = 6 \ "units"$, and $m ang R = 72°$. Using the law of sines and the given information, which of the following is true?
1. No such triangle exists.
2. These measurements result in a unique triangle.
3. These measurements result in two possible triangles.
4. The law of sines cannot be used in this situation.
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing reason in step 10?
1. Substitution Property of Equality
2. Trigonometric ratios in a right triangle
3. Trigonometric Identity
4. Triangle Addition Postulate
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing reason in step 12?
1. Segment Addition Postulate
2. A line segment consists of infinitely many points
3. A line segment can be extended indefinitely as a line
4. There are infinitely many lines in a plane
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
In $Delta LMN$, $LM = 7.5 \ "units"$, $MN = 6 \ "units"$, and $m ang L = 49°$. Using the law of sines and the given information, which of the following is true?
1. No such triangle exists.
2. These measurements result in a unique triangle.
3. These measurements result in two possible triangles.
4. The law of sines cannot be used in this situation.
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing reason in step 16?
3. Substitution Property of Equality
4. Distributive Property
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing reason in step 13?
1. The altitude of an acute angle in an obtuse triangle intersects the extension of the opposite side of the triangle
2. The altitude of a triangle consists of infinitely many points
3. Every triangle has three altitudes
4. The orthocenter of an obtuse triangle lies outside the triangle
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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Is this proof also valid for the other angles in the triangle? Namely, $BC^2 = AB^2 + AC^2 - 2 AB \ AC cos(m ang A)$ and $AC^2 = AB^2 + BC^2 - 2 AB \ BC \ cos(m ang B)$. Why or why not?
1. Yes, since if an altitude was constructed from either remaining vertex, the form of the proof for these other formulas would be identical.
2. Yes, since $m ang A$ and $m ang B$ are both larger than $m ang C$. Therefore the same proof is true for these other formulas.
3. No, since the altitudes from the other vertices may or may not intersect the other sides.
4. No, this proof is only valid for the smallest angle of an acute triangle.
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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What is the missing statement in step 17?
1. $sin(m ang BAE) = (AE)/(AB)$
2. $sin(m ang BAE) = (BE)/(AB)$
3. $sin(m ang BAE) = (BE)/(BC)$
4. $sin(m ang BAE) = (BE)/(AE)$
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