Периферен ъгъл

[Гост]

Здравейте, много се извинявам, че не пиша условията с думи, а прикачвам снимка, но нямам време. Много съжалявам и много моля, помогнете ми!!!!

[Гост]

Много ще съм благодарен, ако ми помогнете!!!

[Гост]

Моля Виииии!!!

[ammornil]

Всеки от ъглите, на който едно от рамената е допирателна към окръжността, а другото рамо пресича окръжността е ПЕРИФЕРЕН.
(оцветените на чертежа)

220228_002.png (9.29 KiB) Прегледано 372 пъти

[ammornil]

(2)

220228_003.png (6.73 KiB) Прегледано 371 пъти

а) [tex]\widehat{ANB}-\widehat{BMA}=180 ^\circ \Rightarrow \widehat{ANB}=\widehat{BMA}+180 ^\circ[/tex]

[tex]\widehat{BMA}+\widehat{BMA}+180 ^\circ=360 ^\circ \Rightarrow 2\widehat{BMA}=180 ^\circ \Rightarrow \widehat{BMA}=90 ^\circ[/tex]

[tex]\angle ABC = \frac{ \widehat{BMA}}{2}=45 ^\circ , \phantom{AA} \angle ABD = 180 ^\circ -\angle ABC=135 ^\circ[/tex]

б) [tex]\widehat{ANB}:\widehat{BMA}=11:7 \Rightarrow \widehat{ANB}=11x, \widehat{BMA}=7x[/tex]

[tex]7x+11x=360 ^\circ \Rightarrow 18x=360 ^\circ \Rightarrow x=20 ^\circ \Rightarrow \begin{cases} \widehat{BMA}=7x=140 ^\circ \\ \widehat{ANB}=11x=220 ^\circ \end{cases}[/tex]

[tex]\angle ABC = \frac{ \widehat{BMA}}{2}=70 ^\circ , \phantom{AA} \angle ABD = 180 ^\circ -\angle ABC=110 ^\circ[/tex]

[ammornil]

(3)

А)

220228_005.png (13.9 KiB) Прегледано 370 пъти

[tex]\angle CAB=40 ^\circ, \phantom{QQ} \angle ACB=110 ^\circ[/tex]
[tex]\widehat{CVB}=2\angle CAB = 80 ^\circ, \phantom{QQ} \widehat{BPA}=2 \angle ACB = 220 ^\circ, \phantom{QQ} \widehat{AQC}=360 ^\circ -(\widehat{CVB}+ \widehat{BPA})= 60 ^\circ[/tex]

[tex]\angle CBL = \frac{\widehat{CVB}}{2}=40 ^\circ[/tex], [tex]\angle ABL = \frac{\widehat{AQC}+\widehat{CVB}}{2}=70 ^\circ[/tex],
[tex]\triangle ABL \rightarrow \angle CAB+\angle ABL+\angle BLA =180 ^\circ \Rightarrow \angle BLA =180 ^\circ-(\angle CAB+\angle ABL)=80 ^\circ[/tex]

Б)

220228_006.png (13.34 KiB) Прегледано 370 пъти

[tex]\angle CAB=3x, \angle CBA=4x, \angle ACB=5x[/tex]
[tex]\angle CAB + \angle CBA+ \angle ACB=180 ^\circ \Rightarrow 12x=180 ^\circ \Rightarrow x=15 ^\circ[/tex]

[tex]\angle CAB=3x=45 ^\circ , \angle CBA=4x=60 ^\circ , \angle ACB=75 ^\circ[/tex]

[tex]\widehat{CVB}=2\angle CAB = 90 ^\circ, \phantom{QQ} \widehat{BPA}=2 \angle ACB = 150 ^\circ, \phantom{QQ} \widehat{AQC}=360 ^\circ -(\widehat{CVB}+ \widehat{BPA})= 120 ^\circ[/tex]

[tex]\angle CBL = \frac{\widehat{CVB}}{2}=45^\circ[/tex], [tex]\angle ABL = \frac{\widehat{AQC}+\widehat{CVB}}{2}=105 ^\circ[/tex],

[tex]\triangle ABL \rightarrow \angle CAB+\angle ABL+\angle BLA =180 ^\circ \Rightarrow \angle BLA =180 ^\circ-(\angle CAB+\angle ABL)=30 ^\circ[/tex]

[Гост]

(3)

А)

220228_005.png

[tex]\angle CAB=40 ^\circ, \phantom{QQ} \angle ACB=110 ^\circ[/tex]
[tex]\widehat{CVB}=2\angle CAB = 80 ^\circ, \phantom{QQ} \widehat{BPA}=2 \angle ACB = 220 ^\circ, \phantom{QQ} \widehat{AQC}=360 ^\circ -(\widehat{CVB}+ \widehat{BPA})= 60 ^\circ[/tex]

[tex]\angle CBL = \frac{\widehat{CVB}}{2}=40 ^\circ[/tex], [tex]\angle ABL = \frac{\widehat{AQC}+\widehat{CVB}}{2}=70 ^\circ[/tex],
[tex]\triangle ABL \rightarrow \angle CAB+\angle ABL+\angle BLA =180 ^\circ \Rightarrow \angle BLA =180 ^\circ-(\angle CAB+\angle ABL)=80 ^\circ[/tex]

Б)

220228_006.png

[tex]\angle CAB=3x, \angle CBA=4x, \angle ACB=5x[/tex]
[tex]\angle CAB + \angle CBA+ \angle ACB=180 ^\circ \Rightarrow 12x=180 ^\circ \Rightarrow x=15 ^\circ[/tex]

[tex]\angle CAB=3x=45 ^\circ , \angle CBA=4x=60 ^\circ , \angle ACB=75 ^\circ[/tex]

[tex]\widehat{CVB}=2\angle CAB = 90 ^\circ, \phantom{QQ} \widehat{BPA}=2 \angle ACB = 150 ^\circ, \phantom{QQ} \widehat{AQC}=360 ^\circ -(\widehat{CVB}+ \widehat{BPA})= 120 ^\circ[/tex]

[tex]\angle CBL = \frac{\widehat{CVB}}{2}=45^\circ[/tex], [tex]\angle ABL = \frac{\widehat{AQC}+\widehat{CVB}}{2}=105 ^\circ[/tex],

[tex]\triangle ABL \rightarrow \angle CAB+\angle ABL+\angle BLA =180 ^\circ \Rightarrow \angle BLA =180 ^\circ-(\angle CAB+\angle ABL)=30 ^\circ[/tex]

МНОГО, МНОГО ВИ БЛАГОДАРЯ!