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Question Number 186737 by Rupesh123 last updated on 09/Feb/23

Commented by Frix last updated on 09/Feb/23

No. There′s a triangle with area (1/(17425)) of the circle.

$$\mathrm{No}. \\ $$$$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{area}\:\frac{\mathrm{1}}{\mathrm{17425}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$

Answered by mr W last updated on 09/Feb/23

Commented by mr W last updated on 09/Feb/23

57+32+25+101+98+47=360 ✓ (q/p)=((sin ((25+101)/2))/(sin ((57+47)/2)))=((sin 63)/(sin 52)) (p/r)=((sin ((57)/2))/(sin ((101)/2)))=((sin 28.5)/(sin 50.5)) (r/q)=((sin ((101+98)/2))/(sin ((32+57)/2)))=((sin 99.5)/(sin 44.5)) (q/p)×(p/r)×(r/q)=((sin 63)/(sin 52))×((sin 28.5)/(sin 50.5))×((sin 99.5)/(sin 44.5)) 1=^(?) ((sin 63 )/(sin 52))((sin 28.5 )/(sin 50.5))((sin 99.5)/(sin 44.5))≈0.98389≠1 ⇒the three lines don′t intersect at the same point!

$$\mathrm{57}+\mathrm{32}+\mathrm{25}+\mathrm{101}+\mathrm{98}+\mathrm{47}=\mathrm{360}\:\checkmark \\ $$$$\frac{{q}}{{p}}=\frac{\mathrm{sin}\:\frac{\mathrm{25}+\mathrm{101}}{\mathrm{2}}}{\mathrm{sin}\:\frac{\mathrm{57}+\mathrm{47}}{\mathrm{2}}}=\frac{\mathrm{sin}\:\mathrm{63}}{\mathrm{sin}\:\mathrm{52}} \\ $$$$\frac{{p}}{{r}}=\frac{\mathrm{sin}\:\frac{\mathrm{57}}{\mathrm{2}}}{\mathrm{sin}\:\frac{\mathrm{101}}{\mathrm{2}}}=\frac{\mathrm{sin}\:\mathrm{28}.\mathrm{5}}{\mathrm{sin}\:\mathrm{50}.\mathrm{5}} \\ $$$$\frac{{r}}{{q}}=\frac{\mathrm{sin}\:\frac{\mathrm{101}+\mathrm{98}}{\mathrm{2}}}{\mathrm{sin}\:\frac{\mathrm{32}+\mathrm{57}}{\mathrm{2}}}=\frac{\mathrm{sin}\:\mathrm{99}.\mathrm{5}}{\mathrm{sin}\:\mathrm{44}.\mathrm{5}} \\ $$$$\frac{{q}}{{p}}×\frac{{p}}{{r}}×\frac{{r}}{{q}}=\frac{\mathrm{sin}\:\mathrm{63}}{\mathrm{sin}\:\mathrm{52}}×\frac{\mathrm{sin}\:\mathrm{28}.\mathrm{5}}{\mathrm{sin}\:\mathrm{50}.\mathrm{5}}×\frac{\mathrm{sin}\:\mathrm{99}.\mathrm{5}}{\mathrm{sin}\:\mathrm{44}.\mathrm{5}} \\ $$$$\mathrm{1}\overset{?} {=}\frac{\mathrm{sin}\:\mathrm{63}\:}{\mathrm{sin}\:\mathrm{52}}\frac{\mathrm{sin}\:\mathrm{28}.\mathrm{5}\:}{\mathrm{sin}\:\mathrm{50}.\mathrm{5}}\frac{\mathrm{sin}\:\mathrm{99}.\mathrm{5}}{\mathrm{sin}\:\mathrm{44}.\mathrm{5}}\approx\mathrm{0}.\mathrm{98389}\neq\mathrm{1} \\ $$$$\Rightarrow{the}\:{three}\:{lines}\:{don}'{t}\:{intersect}\:{at} \\ $$$${the}\:{same}\:{point}! \\ $$

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