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Luda [366]
3 years ago
13

Find the exact value of csc theta if tan theta = sqrt3 and the terminal side of theta is in Quadrant III.

Mathematics
1 answer:
WARRIOR [948]3 years ago
8 0

Answer:

3rd option

Step-by-step explanation:

Using the identities

cot x = \frac{1}{tanx}

csc² x = 1 + cot² x

Given

tanθ = \sqrt{3} , then cotθ = \frac{1}{\sqrt{3} }

csc²θ = 1 + (\frac{1}{\sqrt{3} } )² = 1 + \frac{1}{3} = \frac{4}{3}

cscθ = ± \sqrt{\frac{4}{3} } = ± \frac{2}{\sqrt{3} }

Since θ is in 3rd quadrant, then cscθ < 0

cscθ = - \frac{2}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = - \frac{2\sqrt{3} }{3}

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101 E Victoria Ct, Greenville, NC 27858 my address from mr beast
RideAnS [48]

Answer:

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Step-by-step explanation:

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3 0
2 years ago
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Kelly has 4 dimes and some nickels. The value of her coins is 2.25 how many
Semmy [17]
4 dimes=4 *10=40 cents

2.25-0.40=1.85


1.85=0.05 times number of nickles
times 100 both sides
185=5 times number of nickels
divide both sides by 5
37=number of nickles

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8 0
3 years ago
Which is the simplest radical form of √52?
Zielflug [23.3K]
Short answer: 2 * sqrt(13)
Remark
There are a number of ways to look at this. I'll pick the easiest.

Step one
Factor 52 until there are no more prime factors to be used.

52 = 2 * 26
52 = 2 * 2 * 13. That's as far as you can go.

Rule
For every 2 equal prime factors, 1 of them can be taken out side of the root sign. The other one disappears.

sqrt(52) = sqrt(2*2* 13) = 2*sqrt(13)

Answer 2sqrt(13) <<<< answer
4 0
2 years ago
If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi
Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

BCPO is a square

∴ BC = OP = CP = OB = 210mm

Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

(Hypotenuse)² = Sum of square of the other two sides

(OC)² = (OB)² + (BC)²

(OC)² = 210² + 210²

(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

An isosceless tringle is a triangle that has two equal sides

Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴  ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}

Note that DP = 297 - 210 = 87mm

∠PDO + ∠DOP + 90 = 180

∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

3 0
3 years ago
Which number line represents the solution set for the end equality 3x &lt; -9
shepuryov [24]

divide both sides by 3

so, you'll get x<-3

that's your solution

4 0
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