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1. Answer (D). By the law of sines, we have in any
2. Answer (C). The law of cosines, accepts up to three sides and an angle as an input.
3. Answer (D). Although this triangle is right, we are not given enough information to uniquely determine its sides and angles - here, we need either one more side or one more angle.
4. Answer (D). Don't get tripped up by answer choice (C) - this is just a rearrangement of the statement of the law of cosines. In choice (D), the signs of and
are reversed.
5. Answer (B). By the law of sines, we have Solving gives
Note that this is the <em>ambiguous (SSA) case</em> of the law of sines, where the given measures could specify one triangle, two triangles, or none at all!
6. Answer (A). Since we know all three sides and none of the angles, starting with the law of sines will not help, so we begin with the law of cosines to find one angle; from there, we can use the law of sines to find the remaining angles.
Answer:
<u>Triangle ABC and triangle MNO</u> are congruent. A <u>Rotation</u> is a single rigid transformation that maps the two congruent triangles.
Step-by-step explanation:
ΔABC has vertices at A(12, 8), B(4,8), and C(4, 14).
- length of AB = √[(12-4)² + (8-8)²] = 8
- length of AC = √[(12-4)² + (8-14)²] = 10
- length of CB = √[(4-4)² + (8-14)²] = 6
ΔMNO has vertices at M(4, 16), N(4,8), and O(-2,8).
- length of MN = √[(4-4)² + (16-8)²] = 8
- length of MO = √[(4+2)² + (16-8)²] = 10
- length of NO = √[(4+2)² + (8-8)²] = 6
Therefore:
- AB ≅ MN
- AC ≅ MO
- CB ≅ NO
and ΔABC ≅ ΔMNO by SSS postulate.
In the picture attached, both triangles are shown. It can be seen that counterclockwise rotation of ΔABC around vertex B would map ΔABC into the ΔMNO.
