Math, asked by Gurleen211007, 1 month ago

Need to prove that angle BOC = 90° - 1/2 angle A

Answers

Answered by ItzJanvi10
7

\huge\bold\red{Answer}

ItzJanhvi 10❤

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Answered by deepak1463
8

Step-by-step explanation:

∠1=∠2,∠4=∠5

∠1=∠2,∠4=∠5In ΔABC

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180∠A+2∠2+2∠5=180

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180∠A+2∠2+2∠5=180∠2+∠5=90−21∠A

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180∠A+2∠2+2∠5=180∠2+∠5=90−21∠ANow in ΔBOC

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180∠A+2∠2+2∠5=180∠2+∠5=90−21∠ANow in ΔBOC∠2+∠5+∠BOC=180

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180∠A+2∠2+2∠5=180∠2+∠5=90−21∠ANow in ΔBOC∠2+∠5+∠BOC=18090−21∠A+BOC=180

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180∠A+2∠2+2∠5=180∠2+∠5=90−21∠ANow in ΔBOC∠2+∠5+∠BOC=18090−21∠A+BOC=180∠BOC=90+21∠A

∠1=∠2,∠4=∠5In ΔABC∠A+∠B+∠C=180∠A+2∠2+2∠5=180∠2+∠5=90−21∠ANow in ΔBOC∠2+∠5+∠BOC=18090−21∠A+BOC=180∠BOC=90+21∠A∠BOC=90+21∠BAC

Hope this will help you.

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