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adelina 88 [10]
4 years ago
14

Find csc0 and cos0 , where 0 is the angle shown in the figure. Give exact values, not decimal approximations.

Mathematics
1 answer:
solniwko [45]4 years ago
8 0

Answer:

csc(θ) =  1.1662

cos(θ)  = 0.5145

Step-by-step explanation:

I attach the image of your question in the picture below

We know the values of the adjacent and opposite cathetus

tan(θ) = Opposite / Adjacent

tan(θ) = 5/3 = 1.666

θ = tan^-1 (5/3)

θ = 59.0362 °

Now that we have the angle, we can easily find the value of the expressions needed

csc(θ) = 1 / sin(θ)  = 1/ (0.85749) = 1.1662

cos(θ)  = 0.5145

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Q 1 PLEASE HELP ME FIGURE THIS OUT 2
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Answer: \frac{\pi }{3}

<u>Step-by-step explanation:</u>

cot is \frac{cos}{sin} on the Unit Circle. When is cos = \frac{1}{2} and sin = \frac{\sqrt{3}} {2}?  In other words, where is the coordinate (\frac{1}{2},\frac{\sqrt{3}} {2}) on the Unit Circle?

It occurs at 60° = \frac{\pi }{3} radians

**************************************************************************************

Answer: cot θ

<u>Step-by-step explanation:</u>

  sec (90 - θ) * cos θ

= \frac{1}{cos (90 -\theta)} * cos θ

= \frac{cos\theta}{cos(90 - \theta)}

= \frac{cos\theta}{cos90*cos\theta+sin90*sin\theta}

= \frac{cos\theta}{0*cos\theta+1*sin\theta}

= \frac{cos\theta}{sin\theta}

= cot θ

**************************************************************************************

Answer: BC = √5, AC = 2√5

<u>Step-by-step explanation:</u>

\frac{\pi} {3} = 60°, which means ΔABC is a 30°-60°-90° triangle so we can use the side length formulas: x - x√3 - 2x.

∠C is the 60°. It matches to side AB so: AB = x√3 = √15   ⇒   x = √5

∠A is the 30°. It matches to side BC so: BC = x = √5

∠B is the 90°. It matches to side AC so: AC = 2x = 2(√5)  = 2√5

**************************************************************************************

Answer: k = 4, RS = 12, QS = 28

<u>Step-by-step explanation:</u>

\frac{QR}{MN} = \frac{RS}{NO}= \frac{QS}{MO}

\frac{24}{6} = \frac{RS}{3}= \frac{QS}{7}

proportionality constant k can be found with \frac{QR}{MN} = \frac{24}{6} = 4

\frac{QR}{MN} = \frac{RS}{NO}

→ 4 = \frac{RS}{3}

→ 4(3) = RS

→ 12 = RS

\frac{QR}{MN} = \frac{QS}{MO}

→ 4 = \frac{QS}{7}

→ 4(7) = QS

→ 28 = QS


3 0
3 years ago
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