$\angle BAD = x + 20^\circ$; $\angle ADC = 80^\circ$; $\angle DCB = 100^\circ$; $\angle CBA = 3x$; $x = ?$
- A. $x = 40^{\circ}$
- B. $x = 30^{\circ}$
- C. $x = 50^{\circ}$
- D. $x = 20^{\circ}$
To'rtburchaklar bo‘yicha savol
Javob va yechimni ko‘rish
Javob: $x = 40^{\circ}$
Mavzuni mustahkamlang
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Mavzuni o‘rganishShu mavzudagi boshqa savollar
$\angle BAD = x + 40^\circ$; $\angle ADC = 70^\circ$; $\angle DCB = x + 20^\circ$; $\angle CBA = 3x + 20^\circ$; $x = ?$$\angle FAB = 120^\circ$; $\angle BCE = 70^\circ$; $\angle ABC = 4x$; $\angle CDA = x + 20^\circ$; $x = ?$$\angle FAB = 4x + 20^\circ$; $\angle ABC = 100^\circ$; $\angle BCE = 2x + 10^\circ$; $\angle CDA = 80^\circ$; $x = ?$BG va AG bissektrisa; $\angle AGB = 80^\circ$; $\angle BCD = 120^\circ$; $\angle ADC = x = ?$BG va AG bissektrisa; $\angle AGB = 100^\circ$; $\angle BCD = 2x + 30^\circ$; $\angle CDA = x + 20^\circ$; $x = ?$