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Question-189965




Question Number 189965 by 073 last updated on 25/Mar/23
Answered by som(math1967) last updated on 25/Mar/23
α=32+78=110
$$\alpha=\mathrm{32}+\mathrm{78}=\mathrm{110} \\ $$
Commented by 073 last updated on 25/Mar/23
why?
$$\mathrm{why}? \\ $$
Commented by mnjuly1970 last updated on 25/Mar/23
  C_( 1)  + C_2 =180      C_( 1)  +78= 180
$$\:\:{C}_{\:\mathrm{1}} \:+\:{C}_{\mathrm{2}} =\mathrm{180} \\ $$$$\:\:\:\:{C}_{\:\mathrm{1}} \:+\mathrm{78}=\:\mathrm{180} \\ $$$$\:\:\:\:\:\: \\ $$
Commented by JDamian last updated on 25/Mar/23
Which C? there are 2 points named C in the diagram
Answered by horsebrand11 last updated on 25/Mar/23
∡D+∡B=180°  ∡B=180°−(78°+32°)  ∡D=α=180°−(180°−(110°)))  ∡D=α=110°
$$\measuredangle{D}+\measuredangle{B}=\mathrm{180}° \\ $$$$\measuredangle{B}=\mathrm{180}°−\left(\mathrm{78}°+\mathrm{32}°\right) \\ $$$$\left.\measuredangle{D}=\alpha=\mathrm{180}°−\left(\mathrm{180}°−\left(\mathrm{110}°\right)\right)\right) \\ $$$$\measuredangle{D}=\alpha=\mathrm{110}° \\ $$
Answered by HeferH last updated on 26/Mar/23
Commented by HeferH last updated on 26/Mar/23
ABCD is cyclic⇒∠ DCB = 102°  ∠DCB is external angle of triangle with side  DC ⇒  α =110°
$$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{cyclic}\Rightarrow\angle\:\mathrm{DCB}\:=\:\mathrm{102}° \\ $$$$\angle\mathrm{DCB}\:\mathrm{is}\:\mathrm{external}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{side} \\ $$$$\mathrm{DC}\:\Rightarrow\:\:\alpha\:=\mathrm{110}° \\ $$$$\: \\ $$

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