The solution to the problem is as follows:
<span>cos2A - cos4A = -2 sin(6A/2).sin(-2A/2) = +2 sin(3A).sinA
and
sin4A - sin2A = 2 cos(6A/2).sin(2A/2) = 2 cos(3A).sinA
Hence RHS = (cos(2A)-cos(4A))/ (sin(4A)- Sin(2A)) = sin 3A / cos 3A = tan 3A = LHS
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Answer:
its c
Step-by-step explanation:
i need it on khan academy
<u>Part 1)</u> Find m∠EAD
we know that
m∠EAD+m∠DAF+m∠BAF=
----> by supplementary angles
solve for m∠EAD
m∠EAD=-(m∠DAF+m∠BAF)
in this problem we have
m∠DAF=
m∠BAF=
substitute in the formula above
m∠EAD=
therefore
<u>the answer part 1) is </u>
m∠EAD=
<u>Part 2)</u> Find m∠CAB
we know that
m∠CAB+m∠BAF= --------> by supplementary angles
solve for m∠CAB
m∠CAB=-m∠BAF
in this problem we have
m∠BAF=
substitute in the formula above
m∠CAB=
therefore
<u>the answer part 2) is</u>
m∠CAB=
32/9 as a mises fraction is : 3 5/9
Hope this helps :)
Answer:12 x 9
Step-by-step explanation: 600 x 2/100 = 1200/100 = 12
450 x 2/100 = 900/100 = 9
Remember that 100km = 2 units
