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Since a,b,c form a geometric sequence, then Applying the sine theorem, we have Then or (truncated). Hence Additionally since or

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[Solved] Trigonometry triangles

Triangles & Diagonals

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June 15, 2021 3:28 pm

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June 15, 2021 3:29 pm

Topic starter

Staff

(@armor)

Admin

Honorable Member

317 Posts
156 153 0

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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