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Charra [1.4K]
3 years ago
13

Help me please

Mathematics
1 answer:
kari74 [83]3 years ago
6 0

Answer:

\frac{ \cos(80) }{ \sin(10) }  +  \frac{ \sin(20) }{ \cos(70) }  = 2 \\  \frac{ \cos(90 - 10) }{ \sin(10) }  +  \frac{ \sin(20) }{ \cos(9 0 - 20) }  = 2 \\  \frac{ \sin(10) }{ \sin(10) }  +  \frac{ \sin(20) }{ \sin(20) }  = 2 \\ 1 + 1 = 2 \\ 2 = 2 \\  \\  \frac{ \cot(40) }{ \tan(50) }  +  \frac{ \cos(65) }{ \sin(115) }  = 2 \\  \frac{ \cot(90 - 50) }{ \tan(50) }  +  \frac{ \cos(65) }{ \sin(90 + 65) }  = 2 \\  \frac{ \tan(50) }{ \tan(50) }  +  \frac{ \cos(65) }{ \cos(65) }  = 2 \\ 1 + 1 \\  = 2

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Determine the unknown angle measures in the figure.<br><br>HELP NOW!!! please
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Answer:

m∠1=80°

m∠2=112°

m∠3=131°

m∠4=80°

m∠5=37°

Step-by-step explanation:

First you have to find m∠2

To do that find m∠6 (I created this angle shown in pic below)

Find m∠6 by using the sum of all ∠'s in a Δ theorem

m∠6=180°-(63°+49°)

m∠6=68°

Now you can find m∠2 with the supplementary ∠'s theorem

m∠2=180°-68°

m∠2=112°

Then you find m∠5 using the sum of all ∠'s in a Δ theorem

m∠5=180°-(112°+31°)

m∠5=37°

Now you can find m∠1

m∠1=180°-(63°+37°)

m∠1=180°-100°=80°

m∠4 can easily be found too now:

m∠4=180°-(63°+37°)

m∠4=80°

m∠3=180°-49°

m∠3=131°

8 0
3 years ago
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