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zysi [14]
3 years ago
9

What are the measures of ∠1 and ∠2? (HELP ASAP USE IMAGE!!)

Mathematics
2 answers:
koban [17]3 years ago
7 0
Answer letter b I think
miv72 [106K]3 years ago
4 0
Not sure i thing it’s C or B. sorry if i’m wrong! have a good day:)
You might be interested in
Use distributive property to rewrite 9(7 8)
777dan777 [17]
9(7 + 8) = (9 x 7) + (9 x 8) = 63 + 72 = 135
8 0
3 years ago
If p is the incenter of triangle jkl, find each measure
Hunter-Best [27]

Answer:

PO = 7

PM = 7

MJ = 11

∠PJO = 32°

∠KJL = 64°

PL ≈ 18.385

OL = 17

∠PLO = 22°

∠NLO = 44°

∠JKL = 72°

∠MKP = 36°

∠NKP = 36°

∠PKN = 36°

KN = 10

PL ≈ 13.04

PK ≈ 12.21

JL = 28

JK = 21

LK = 27

Step-by-step explanation:

The given parameters are;

The point representing the incenter of the triangle = P

Therefore PO = PM = PN = 7

tan(32°) = PM/JM = 7/JM

∴JM = 7/(tan(32°)) ≈ 11.2

∠PJO = tan⁻¹(7/11)≈ 32.47°

∠PJO = ∠PJM = 32° similar triangles

∠KJL = ∠KJP + ∠PJO = 32 + 32 = 64°

∠KJL ≈ 64°

PL = √(7² + 17²) ≈ 18.385

OL = NL = 17 similar triangles

∠PLO = sin⁻¹(7/18.385) ≈ 22.380°

∠PLO = ∠PLN = 22°

∠NLO = ∠PNL + ∠OLP ≈ 22° + 22° ≈ 44°

∠NLO ≈ 44.380°

∠JKL = 180 - (∠KJL + ∠NLO)

∠JKL = 180° - (64° + 44°) ≈ 72°

∠JKL  ≈ 72°

∠MKP = ∠NKP = 72°/2 = 36°

∠MKP = 36°

∠NKP = 36°

∠PKN = ∠JKL - ∠MKP = 72° - 36° ≈ 36°

∠PKN  ≈ 36°

KN = KM = 10

MJ = OJ = 11

PL = √(7² + 11²) ≈ 13.04

PK = √(7² + 10²) ≈ 12.21

JL = JO + OL = 11 + 17 = 28

JK = JM + MK = 11 + 10 = 21

LK = LN + NK = 17 + 10 = 27

6 0
3 years ago
Can someone help with these trigonometry problems
bonufazy [111]
4. What you want to do is first find the tan ratio, because tangent is opposite / adjacent, which you can use to find x.

So tan(52) = 1.28 ish. So we know that x/18 = 1.28.
Then multiply both sides by 18.

x / 18 * 18 = 1.28 * 18

x = about 23.

5. Now, you can use the pythagorean's theorem, a^2 + b^2 = c^2.

Just input all values.

18 ^ 2 + 23 ^ 2 = c ^ 2

324 + 529 = c ^ 2

853 = c^2

square root of 853 = c

x = 29.2 ish

6. In this case, use cosine, or adjacent / hypotenuse.

cos(16) = 0.96 ish

so that means 24/x = 0.96

Then basically,

24 / 0.96 = x

x = 25

So x = 25

Then you can use the Pythagorean's Theorem again. I'm not going to go as in depth.

a ^ 2 + b ^ 2 = c ^ 2

24 ^ 2 + b ^ 2 = 25 ^ 2

576 + b ^ 2 = 625

b ^ 2 = 49

b = 7

8. Use tan here. opposite/adjacent Again, not going as in depth.

tan(55) = 1.43 ish

1.43 = Y / 22

1.42 * 22 = Y

Y = 31.24 ish

Length of MN:

31.24 ^ 2 + 22 ^ 2 = c ^ 2

976ish + 484 = c ^ 2

1460 = c ^ 2

x = 38.2 ish

10. Use cosine.

cos(63) = about 0.45

Adjacent / hypotenuse

3 / x = 0.45

3 / 0.45= x

x = 6.67

And then pythagorean's Theorem.

6.67 ^ 2 = 3 ^ 2 + b ^ 2

44.5 = 9 + b ^ 2

35.5 = b^2

b = about 6
4 0
3 years ago
|3(x-4)|+7=16. this is my question but i am not sure how to solve it for two values of x.
denis-greek [22]

Answer:

7

Step-by-step explanation:

(3x-12)+7=16

3x-12=16-7

3x-12=9

3x=9+12

3x=21

x=21:3

x=7

7 0
3 years ago
StartFraction 2 Over n EndFraction = two-thirds
KiRa [710]

Answer:

3 Kfc

Step-by-step explanation:

2/3 times 9/2 is 18/6 which gives you 3

7 0
3 years ago
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