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[Solved] Trigonometry triangles

 
Triangles & Diagonals
Last Post by Staff 4 years ago
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June 15, 2021 4:28 pm
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(@armor)
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June 15, 2021 4:29 pm
Topic starter
Staff 
(@armor)
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Since a,b,c form a geometric sequence, then \small b^2=ac.  Applying the sine theorem, we have \small sin^2B=sinAsinC.  Then 

\small 1-cos(A-C)\Rightarrow 2cos^2B+cosB-1\geq 0\Rightarrow cosB\geq \frac{1}{2}, 

or    \small cosB\leq -1  (truncated).

Hence \small 0<B\leq \frac{\Pi }{3}.

Additionally since   \small sinB+cosB=\sqrt{2}sin(B+\frac{\Pi }{4})\Rightarrow 1\leq m^2\leq \sqrt{2}  

\small \Rightarrow -\sqrt[4]{2}\leq m\leq -1,   or  \small 1\leq m\leq \sqrt[4]{2}.

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