Question-148090

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Question Number 148090 by mathdanisur last updated on 25/Jul/21

Commented by Rasheed.Sindhi last updated on 25/Jul/21

$$\angle{B}=\mathrm{75}°×\mathrm{2}=\mathrm{150}° \\ $$

Commented by mathdanisur last updated on 25/Jul/21

$${Thank}\:{you}\:{Ser},\:{but}\:{how}.? \\ $$

Answered by Rasheed.Sindhi last updated on 25/Jul/21

Join O to B & C △AOB: ∵OA=OB [radii of same circle] ∴ ∠A=∠ABO=75° [opposite angles of conguent sides] △ABO↔△CBO AB=CB (given) AO=CO [radii of same circle] BO=BO [common side] ∴△ABO≅△CBO [SSS≅SSS] ∴∠ABO=∠CBO=75°[corresponding angles of congruent triangles] ∠ABC=∠ABO+∠CBO =75°+75°=150° ∠B=150°

$${Join}\:{O}\:{to}\:{B}\:\&\:{C} \\ $$$$\bigtriangleup{AOB}: \\ $$$$\because{OA}={OB}\:\:\left[{radii}\:{of}\:{same}\:{circle}\right] \\ $$$$\therefore\:\angle{A}=\angle{ABO}=\mathrm{75}°\:\left[{opposite}\:{angles}\:{of}\:{conguent}\:{sides}\right] \\ $$$$\bigtriangleup{ABO}\leftrightarrow\bigtriangleup{CBO} \\ $$$$\:\:\:\:\:{AB}={CB}\:\left({given}\right) \\ $$$$\:\:\:\:\:{AO}={CO}\:\:\left[{radii}\:{of}\:{same}\:{circle}\right] \\ $$$$\:\:\:\:\:\:{BO}={BO}\:\left[{common}\:{side}\right] \\ $$$$\therefore\bigtriangleup{ABO}\cong\bigtriangleup{CBO}\:\left[{SSS}\cong{SSS}\right] \\ $$$$\:\:\therefore\angle{ABO}=\angle{CBO}=\mathrm{75}°\left[{corresponding}\:{angles}\:{of}\:{congruent}\:{triangles}\right] \\ $$$$\:\:\:\:\:\angle{ABC}=\angle{ABO}+\angle{CBO} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{75}°+\mathrm{75}°=\mathrm{150}° \\ $$$$\:\:\:\:\:\:\:\:\angle{B}=\mathrm{150}° \\ $$

Commented by mathdanisur last updated on 25/Jul/21