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Question Number 177975 by aurpeyz last updated on 11/Oct/22

Answered by mr W last updated on 11/Oct/22

∠ACB=x/2 ∠BAC=y ∠B=180−x/2−y ∠ADC=180−∠B=x/2+y

$$\angle{ACB}={x}/\mathrm{2} \\ $$$$\angle{BAC}={y} \\ $$$$\angle{B}=\mathrm{180}−{x}/\mathrm{2}−{y} \\ $$$$\angle{ADC}=\mathrm{180}−\angle{B}={x}/\mathrm{2}+{y} \\ $$

Commented by Tawa11 last updated on 11/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by aurpeyz last updated on 11/Oct/22

pls how is ∠ACB=x/2. thats the only part that i didnt get. thanks

$${pls}\:{how}\:{is}\:\angle{ACB}={x}/\mathrm{2}.\:{thats}\:{the}\:{only}\:{part} \\ $$$${that}\:{i}\:{didnt}\:{get}.\:{thanks} \\ $$

Commented by mr W last updated on 12/Oct/22

Commented by aurpeyz last updated on 12/Oct/22

thanks sir

$${thanks}\:{sir} \\ $$

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