Name one pair of rays that are not opposite rays.
1 answer:
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Answer:
<u>Figure A</u>
Step-by-step explanation:
See the attached figure which represents the given options
We are to select the correct pair of triangles that can be mapped to each other using a translation and a rotation about point A.
As shown: point A will map to point L, point R will map to point P and point Q will map to point K.
we will check the options:
<u>Figure A</u>: the triangle ARQ and LPK can be mapped to each other using a translation and a rotation about point A.
<u>Figure B: </u> the triangle ARQ and LPK can be mapped to each other using a translation and a reflection about the line RA.
<u>Figure C:</u> the triangle ARQ and LPK can be mapped to each other using a translation and a reflection about the line QA.
<u>Figure D:</u> the triangle ARQ and LPK can be mapped to each other using a rotation about point A.
So, the answer is figure A
<u>The triangle pairs of figure A can be mapped to each other using a translation and a rotation about point A.</u>
Very nice problem. You should look up the history of this question. It has a very long one having to do with mirrors made between 50 BC and 50 AD.
Step One
Find AB, BC, AC
Before beginning this problem, I'm going to arbitrarily name
c = AB
a = BC
b = AC It just makes the calculations easier.
AB = radius of the large circle - radius of the medium circle = 20.62 - 8.04 = 12.58
BC = Radii of the two smaller circles added together = 8.04 + 7.35 = 15.39
AC = Radius of the large circle - Radius of the small circle = 20.62 - 7.35 = 13.27
So
a = 15.39
b = 13.27
c = 12.58
Step Two
Find Angle A or <BAC
a^2 = b^2 + c^2 - 2*b*c * Cos(A)
15.39^2 = 12.58^2 + 13.27^2 - 2 * 12.58 * 13.27 * Cos(A)
236.852 = 158.26 + 176.1 - 333.87 * Cos(A)
236.852 = 334.4 - 333.87*Cos(A)
-97.55 = - 333.87 * Cos(A)
-97.55 / -333.87 = Cos(A)
0.2922 = cos(A)
A = cos-1(0.2922)
A = 73.01 degrees
Step Three
Just to see if you understand what was done, I'll give you the givens for finding c, and the answer and you can work through the calculations to see if your answer agrees with mine. If it doesn't PM me.
c = 12.58
b = 13.27
a = 15.39
12.58^2 = 13.27^2 + 15.39^2 - 2*13.27*15.39*Cos(C)
C = 51.42
Step Four
Find <B
Every triangle has 180 degrees so
B = 180 - <A - C
B = 180 - 51.42 - 74.01
B = <54.57.
I have found a moderator who opened the question up so that I can show you why the Cos law is the only way to do this. If the circles are very disproportionate as in this diagram, then no simple assumption can be made. The cos law is all that will work. I would have posted this earlier, but I didn't think anyone would find another method. It's ingenious but not possible for the situation below.
Step One
Find AB, BC, AC
Before beginning this problem, I'm going to arbitrarily name
c = AB
a = BC
b = AC It just makes the calculations easier.
AB = radius of the large circle - radius of the medium circle = 20.62 - 8.04 = 12.58
BC = Radii of the two smaller circles added together = 8.04 + 7.35 = 15.39
AC = Radius of the large circle - Radius of the small circle = 20.62 - 7.35 = 13.27
So
a = 15.39
b = 13.27
c = 12.58
Step Two
Find Angle A or <BAC
a^2 = b^2 + c^2 - 2*b*c * Cos(A)
15.39^2 = 12.58^2 + 13.27^2 - 2 * 12.58 * 13.27 * Cos(A)
236.852 = 158.26 + 176.1 - 333.87 * Cos(A)
236.852 = 334.4 - 333.87*Cos(A)
-97.55 = - 333.87 * Cos(A)
-97.55 / -333.87 = Cos(A)
0.2922 = cos(A)
A = cos-1(0.2922)
A = 73.01 degrees
Step Three
Just to see if you understand what was done, I'll give you the givens for finding c, and the answer and you can work through the calculations to see if your answer agrees with mine. If it doesn't PM me.
c = 12.58
b = 13.27
a = 15.39
12.58^2 = 13.27^2 + 15.39^2 - 2*13.27*15.39*Cos(C)
C = 51.42
Step Four
Find <B
Every triangle has 180 degrees so
B = 180 - <A - C
B = 180 - 51.42 - 74.01
B = <54.57.
I have found a moderator who opened the question up so that I can show you why the Cos law is the only way to do this. If the circles are very disproportionate as in this diagram, then no simple assumption can be made. The cos law is all that will work. I would have posted this earlier, but I didn't think anyone would find another method. It's ingenious but not possible for the situation below.