I need help with #5
2 answers:
Step-by-step explanation:
Well first we can say measure angle 2 equals 130° because they are S.s.int.ext angles, (same side interior angles and exterior). Or also corresponding
Now with measure angle 1, it is a supplementary angle
Let 1 stand for measure angle 1
Subtract 130° from both sides
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Answer: Camera 3 covers the largest angle
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Explanation:
Refer to the diagram shown in the attachment
Let
A = position of camera 1
B = position of camera 2
C = position of camera 3
We need to find the angle values for A, B and C
Notice how
a = distance from camera 2 to camera 3 (B to C)
a = 122
the little 'a' is opposite the big 'A'
Also notice that
b = distance from camera 1 to camera 3 (A to C)
b = 139
the little b is opposite the big B
Also notice that
c = distance from camera 1 to camera 2 (A to B)
c = 151
the little c is opposite the big C
So in summary so far we have
a = 122
b = 139
c = 151
we will use these values in the sections below
Make sure your calculator is in degree mode
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Use the law of cosines to find the measure of angle A
a^2 = b^2 + c^2 - 2*b*c*cos(A)
122^2 = 139^2 + 151^2 - 2*139*151*cos(A)
14884 = 19321 + 22801 - 41978*cos(A)
14884 = 42122 - 41978*cos(A)
14884 - 42122 = 42122-41978*cos(A)-42122
-27238 = -41978*cos(A)
-41978*cos(A) = -27238
-41978*cos(A)/(-41978) = (-27238)/(-41978)
cos(A) = 0.64886369050455
arccos(cos(A)) = arccos(0.64886369050455)
A = 49.544016303819
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Use law of cosines to find angle B
b^2 = a^2 + c^2 - 2*a*c*cos(B)
139^2 = 122^2 + 151^2 - 2*122*151*cos(B)
19321 = 14884 + 22801 - 36844*cos(B)
19321 = 37685 - 36844*cos(B)
19321 - 37685 = 37685-36844*cos(B)-37685
-18364 = -36844*cos(B)
-36844*cos(B) = -18364
-36844*cos(B)/(-36844) = (-18364)/(-36844)
cos(B) = 0.498425795244816
arccos(cos(B)) = arccos(0.498425795244816)
B = 60.1040940123789
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Use law of cosines to find angle C
c^2 = a^2 + b^2 - 2*a*b*cos(C)
151^2 = 122^2 + 139^2 - 2*122*139*cos(C)
22801 = 14884 + 19321 - 33916*cos(C)
22801 = 34205 - 33916*cos(C)
22801 - 34205 = 34205-33916*cos(C)-34205
-11404 = -33916*cos(C)
-33916*cos(C) = -11404
-33916*cos(C)/(-33916) = (-11404)/(-33916)
cos(C) = 0.336242481424696
arccos(cos(C)) = arccos(0.336242481424696)
C = 70.3518896838022
-----------------------------
Or we can use a shortcut to find angle C
A+B+C = 180
C = 180-A-B
C = 180-49.544016303819-60.1040940123789
C = 70.3518896838022
-----------------------------
Summary:
angle A = 49.544016303819 degrees
angle B = 60.1040940123789 degrees
angle C = 70.3518896838022 degrees
Angle C is the largest so camera 3 sweeps out the largest angle
---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
Explanation:
Refer to the diagram shown in the attachment
Let
A = position of camera 1
B = position of camera 2
C = position of camera 3
We need to find the angle values for A, B and C
Notice how
a = distance from camera 2 to camera 3 (B to C)
a = 122
the little 'a' is opposite the big 'A'
Also notice that
b = distance from camera 1 to camera 3 (A to C)
b = 139
the little b is opposite the big B
Also notice that
c = distance from camera 1 to camera 2 (A to B)
c = 151
the little c is opposite the big C
So in summary so far we have
a = 122
b = 139
c = 151
we will use these values in the sections below
Make sure your calculator is in degree mode
----------------------------------
Use the law of cosines to find the measure of angle A
a^2 = b^2 + c^2 - 2*b*c*cos(A)
122^2 = 139^2 + 151^2 - 2*139*151*cos(A)
14884 = 19321 + 22801 - 41978*cos(A)
14884 = 42122 - 41978*cos(A)
14884 - 42122 = 42122-41978*cos(A)-42122
-27238 = -41978*cos(A)
-41978*cos(A) = -27238
-41978*cos(A)/(-41978) = (-27238)/(-41978)
cos(A) = 0.64886369050455
arccos(cos(A)) = arccos(0.64886369050455)
A = 49.544016303819
-----------------------------
Use law of cosines to find angle B
b^2 = a^2 + c^2 - 2*a*c*cos(B)
139^2 = 122^2 + 151^2 - 2*122*151*cos(B)
19321 = 14884 + 22801 - 36844*cos(B)
19321 = 37685 - 36844*cos(B)
19321 - 37685 = 37685-36844*cos(B)-37685
-18364 = -36844*cos(B)
-36844*cos(B) = -18364
-36844*cos(B)/(-36844) = (-18364)/(-36844)
cos(B) = 0.498425795244816
arccos(cos(B)) = arccos(0.498425795244816)
B = 60.1040940123789
-----------------------------
Use law of cosines to find angle C
c^2 = a^2 + b^2 - 2*a*b*cos(C)
151^2 = 122^2 + 139^2 - 2*122*139*cos(C)
22801 = 14884 + 19321 - 33916*cos(C)
22801 = 34205 - 33916*cos(C)
22801 - 34205 = 34205-33916*cos(C)-34205
-11404 = -33916*cos(C)
-33916*cos(C) = -11404
-33916*cos(C)/(-33916) = (-11404)/(-33916)
cos(C) = 0.336242481424696
arccos(cos(C)) = arccos(0.336242481424696)
C = 70.3518896838022
-----------------------------
Or we can use a shortcut to find angle C
A+B+C = 180
C = 180-A-B
C = 180-49.544016303819-60.1040940123789
C = 70.3518896838022
-----------------------------
Summary:
angle A = 49.544016303819 degrees
angle B = 60.1040940123789 degrees
angle C = 70.3518896838022 degrees
Angle C is the largest so camera 3 sweeps out the largest angle