Question

Q 1 PLEASE HELP ME FIGURE THIS OUT 2

Answer 1

Answer: $\frac{\pi }{3}$

Step-by-step explanation:

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

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Answer: cot θ

Step-by-step explanation:

  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

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Answer: BC = √5, AC = 2√5

Step-by-step explanation:

$\frac{\pi} {3}$ = 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

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

Step-by-step explanation:

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

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

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