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[Solved] Geometric sequence trigonometry

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 3:07 pm
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Staff 
(@armor)
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June 15, 2021 3:08 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

(1)  

\small cosB=\frac{3}{4}\Rightarrow 0< B< \frac{\Pi }{2}\Rightarrow sinB=\sqrt{1-(\frac{3}{4})^2}=\frac{\sqrt{7}}{4}.

Since a,b,c form a geometric sequence, applying the sine theorem, we have \small sin^2B=sinA.sinC.

Therefore

\small cotA+cotC=\frac{cosA}{sinA}+\frac{cosC}{sinC}=\frac{sin(A+C)}{sinAsinC}=\frac{sinB}{sin^2B}=\frac{1}{sinB}=\frac{4\sqrt{7}}{7}

(2)

\small \overrightarrow{BA}\overrightarrow{BC}=\frac{3}{2}\Rightarrow cacosB=\frac{3}{2}.  Additionally \small cosB=\frac{3}{4},    thus  \small ca=2.

Since \small b^2=ac=2,  applying the cosine theorem  \small b^2=a^2+c^2-2accosB,   we have

\small a^2+c^2=b^2+2accosB=5\Rightarrow (a+c)^2=a^2+c^2+2ac=5+4=9.  Therefore, \small a+c=3.

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