Answer: [tex]\frac{\pi }{3}[/tex]
Step-by-step explanation:
cot is [tex]\frac{cos}{sin}[/tex] on the Unit Circle. When is cos = [tex]\frac{1}{2}[/tex] and sin = [tex]\frac{\sqrt{3}} {2}[/tex]? In other words, where is the coordinate [tex](\frac{1}{2},\frac{\sqrt{3}} {2})[/tex] on the Unit Circle?
It occurs at 60° = [tex]\frac{\pi }{3}[/tex] radians
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Answer: cot θ
Step-by-step explanation:
sec (90 - θ) * cos θ
= [tex]\frac{1}{cos (90 -\theta)}[/tex] * cos θ
= [tex]\frac{cos\theta}{cos(90 - \theta)}[/tex]
= [tex]\frac{cos\theta}{cos90*cos\theta+sin90*sin\theta}[/tex]
= [tex]\frac{cos\theta}{0*cos\theta+1*sin\theta}[/tex]
= [tex]\frac{cos\theta}{sin\theta}[/tex]
= cot θ
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Answer: BC = √5, AC = 2√5
Step-by-step explanation:
[tex]\frac{\pi} {3}[/tex] = 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:
[tex]\frac{QR}{MN} = \frac{RS}{NO}= \frac{QS}{MO}[/tex]
[tex]\frac{24}{6} = \frac{RS}{3}= \frac{QS}{7}[/tex]
proportionality constant k can be found with [tex]\frac{QR}{MN}[/tex] = [tex]\frac{24}{6}[/tex] = 4
[tex]\frac{QR}{MN} = \frac{RS}{NO}[/tex]
→ 4 = [tex]\frac{RS}{3}[/tex]
→ 4(3) = RS
→ 12 = RS
[tex]\frac{QR}{MN} = \frac{QS}{MO}[/tex]
→ 4 = [tex]\frac{QS}{7}[/tex]
→ 4(7) = QS
→ 28 = QS
