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Question-92635

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Question Number 92635 by otchereabdullai@gmail.com last updated on 08/May/20
Commented by mr W last updated on 08/May/20
∠BFD=(( ∠BOD)/2)=((120)/2)=60° ∠FBC=∠OBC−∠OBG=90−((30)/2)=75° ∠ODB=((180−120)/2)=30° ∠BCD=180−120=60°
$$\angle{BFD}=\frac{\:\:\angle{BOD}}{\mathrm{2}}=\frac{\mathrm{120}}{\mathrm{2}}=\mathrm{60}° \\ $$$$\angle{FBC}=\angle{OBC}−\angle{OBG}=\mathrm{90}−\frac{\mathrm{30}}{\mathrm{2}}=\mathrm{75}° \\ $$$$\angle{ODB}=\frac{\mathrm{180}−\mathrm{120}}{\mathrm{2}}=\mathrm{30}° \\ $$$$\angle{BCD}=\mathrm{180}−\mathrm{120}=\mathrm{60}° \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/May/20
Thank you alot prof w
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{alot}\:\mathrm{prof}\:\mathrm{w} \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/May/20
I respect you a lot Prof W thanks for your time
$$\mathrm{I}\:\mathrm{respect}\:\mathrm{you}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{Prof}\:\mathrm{W} \\ $$$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time} \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/May/20
You are the most powerful professor in mathematics i salute you once again prof w
$$\mathrm{You}\:\mathrm{are}\:\mathrm{the}\:\mathrm{most}\:\mathrm{powerful}\:\mathrm{professor}\:\mathrm{in}\: \\ $$$$\mathrm{mathematics}\:\mathrm{i}\:\mathrm{salute}\:\mathrm{you}\:\mathrm{once}\:\mathrm{again} \\ $$$$\mathrm{prof}\:\mathrm{w} \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/May/20
prof please i have only quesion to understand why is BCD=180−120?
$$\mathrm{prof}\:\mathrm{please}\:\mathrm{i}\:\mathrm{have}\:\mathrm{only}\:\mathrm{quesion}\:\mathrm{to}\: \\ $$$$\mathrm{understand} \\ $$$$\mathrm{why}\:\mathrm{is}\:\mathrm{BCD}=\mathrm{180}−\mathrm{120}? \\ $$$$ \\ $$
Commented by mr W last updated on 08/May/20
Σall angles of a quadrilateral=2×180° ∠BCD+∠BOD+∠OBC+∠ODC=2×180° ∠OBC=∠ODC=90° ⇒∠BCD+∠BOD=180° ⇒∠BCD=180°−∠BOD=180°−120°=60°
$$\Sigma{all}\:{angles}\:{of}\:{a}\:{quadrilateral}=\mathrm{2}×\mathrm{180}° \\ $$$$\angle{BCD}+\angle{BOD}+\angle{OBC}+\angle{ODC}=\mathrm{2}×\mathrm{180}° \\ $$$$\angle{OBC}=\angle{ODC}=\mathrm{90}° \\ $$$$\Rightarrow\angle{BCD}+\angle{BOD}=\mathrm{180}° \\ $$$$\Rightarrow\angle{BCD}=\mathrm{180}°−\angle{BOD}=\mathrm{180}°−\mathrm{120}°=\mathrm{60}° \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/May/20
God please add more years to this man′s age he is helping us alot
$$\mathrm{God}\:\mathrm{please}\:\mathrm{add}\:\mathrm{more}\:\mathrm{years}\:\mathrm{to}\:\mathrm{this}\: \\ $$$$\mathrm{man}'\mathrm{s}\:\mathrm{age}\:\mathrm{he}\:\mathrm{is}\:\mathrm{helping}\:\mathrm{us}\:\mathrm{alot} \\ $$
Commented by mr W last updated on 08/May/20
Commented by I want to learn more last updated on 08/May/20
Amin
$$\mathrm{Amin} \\ $$

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