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Lynna [10]
3 years ago
13

Sin10+sin20+sin40+sin50=sin70+sin80.prove it

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
4 0
We are going to prove it like this:
Lets use the formula sinA+sinB=2sin(A+B/2)cos(A-B/2)
 Now we are going to take the left side of equation
sin10+sin40+sin50+sin20
 Arranging =(sin50+sin10)+(sin40+sin20)
Applying the above formula. =2sin(50+10/2)cos(50-10/2)+2sin(40+20/… =2sin(30)cos(20)+2sin(30)cos(10)
=2sin30{cos20+cos10}
Again using the formula
cosA+cosB= 2cos(A+B/2)cos(A-B/2)
=2sin30{2cos(20+10/2).cos(20-10/2)}
=2sin30{2cos(15).cos(5)}
=2(1/2){2cos15.cos5} as sin30=1/2
=2cos15.cos5
Taking right side of equation sin70+ sin80
Using the formula sinA+sinB
= 2sin(A+B/2)cos(A-B/2)
=2sin(70+80/2)cos(70-80/2)
=2sin75cos5
=2sin(90-15)cos5
=2cos15.cos5 <span>Hope this helps</span>
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3 years ago
Can someone help me
kolbaska11 [484]

m < 1 = 130°

m < 1 = 130°m < 2 = 50°

We know this is a linear pair, therefore the two angles must add up to 180°

(m < 1 )+ (m < 2 )= 180

5x + x + 24 = 180

Combine like terms

6x + 24 = 180

Isolate the variable

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Plug in 26 for x in both of the individual angle equations.

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= 5(26)

= 130

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= (26) + 24

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A good way to check if we did this correctly, is to add up the two angles. Because 180 + 50 = 180, we did this correctly.

I hope this helped. Let me know if you have any questions!

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