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Прилагам снимка със задачата!

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Без заглавие - 2022-03-01T161303.199.png (148.03 KiB) Прегледано 904 пъти

[tex]CH \bot AB , H \in AB[/tex]
[tex]AB = c \Rightarrow AH = HB = \frac{c}{2} , \angle ACB = \gamma \Rightarrow \angle HCB = \frac{ \gamma }{2} ,CH = 2 dm = 20 cm[/tex]
От [tex]\triangle HBC \rightarrow \frac{HB}{HC} = \tg \frac{ \gamma }{2} \Leftrightarrow \frac{c}{2} = 20\tg \frac{ \gamma }{2} \Rightarrow c = 40\tg \frac{ \gamma }{2}[/tex]
За [tex]\triangle ABC[/tex] прилагам Синусова теорема:

[tex]\frac{AB}{\sin \gamma } = 2R \Leftrightarrow c = 60\sin \gamma \Leftrightarrow 40\tg \frac{ \gamma }{2} = 60\sin \gamma \Leftrightarrow 2 \frac{\sin \frac{ \gamma }{2} }{\cos \frac{ \gamma }{2} } = 3.2\sin \frac{ \gamma }{2}\cos \frac{ \gamma }{2} \Rightarrow[/tex]
[tex]\cos^{2 } \frac{ \gamma }{2} = \frac{1}{3} , \sin^{2 } \frac{ \gamma }{2} = \frac{2}{3}[/tex]
[tex]\Rightarrow \sin \frac{ \gamma }{2} = \frac{ \sqrt{2} }{ \sqrt{3} }, \cos \frac{ \gamma }{2} = \frac{1}{ \sqrt{3} }[/tex]
[tex]\sin \gamma = 2.\sin \frac{ \gamma }{2}.\cos \frac{ \gamma }{2} = 2. \frac{ \sqrt{2} }{ \sqrt{3} }. \frac{1}{ \sqrt{3} } = \frac{2 \sqrt{2} }{3}[/tex] - отговор б)