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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

$$\mathrm{\angle }BFD=\frac{\mathrm{\angle }BOD}{2}=\frac{120}{2}=60°$$$$\mathrm{\angle }FBC=\mathrm{\angle }OBC-\mathrm{\angle }OBG=90-\frac{30}{2}=75°$$$$\mathrm{\angle }ODB=\frac{180-120}{2}=30°$$$$\mathrm{\angle }BCD=180-120=60°$$

Commented by otchereabdullai@gmail.com last updated on 08/May/20

$$\mathrm{Thank}\mathrm{you}\mathrm{alot}\mathrm{prof}\mathrm{w}$$

Commented by otchereabdullai@gmail.com last updated on 08/May/20

$$\mathrm{I}\mathrm{respect}\mathrm{you}\mathrm{a}\mathrm{lot}\mathrm{Prof}\mathrm{W}$$$$\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{time}$$

Commented by otchereabdullai@gmail.com last updated on 08/May/20

$$\mathrm{You}\mathrm{are}\mathrm{the}\mathrm{most}\mathrm{powerful}\mathrm{professor}\mathrm{in}$$$$\mathrm{mathematics}\mathrm{i}\mathrm{salute}\mathrm{you}\mathrm{once}\mathrm{again}$$$$\mathrm{prof}\mathrm{w}$$

Commented by otchereabdullai@gmail.com last updated on 08/May/20

$$\mathrm{prof}\mathrm{please}\mathrm{i}\mathrm{have}\mathrm{only}\mathrm{quesion}\mathrm{to}$$$$\mathrm{understand}$$$$\mathrm{why}\mathrm{is}\mathrm{BCD}=180-120?$$$$$$

Commented by mr W last updated on 08/May/20

$$\mathrm{\Sigma }allanglesofaquadrilateral=2\times 180°$$$$\mathrm{\angle }BCD+\mathrm{\angle }BOD+\mathrm{\angle }OBC+\mathrm{\angle }ODC=2\times 180°$$$$\mathrm{\angle }OBC=\mathrm{\angle }ODC=90°$$$$\Rightarrow \mathrm{\angle }BCD+\mathrm{\angle }BOD=180°$$$$\Rightarrow \mathrm{\angle }BCD=180°-\mathrm{\angle }BOD=180°-120°=60°$$

Commented by otchereabdullai@gmail.com last updated on 08/May/20

$$\mathrm{God}\mathrm{please}\mathrm{add}\mathrm{more}\mathrm{years}\mathrm{to}\mathrm{this}$$$${\mathrm{man}}^{\prime }\mathrm{s}\mathrm{age}\mathrm{he}\mathrm{is}\mathrm{helping}\mathrm{us}\mathrm{alot}$$

Commented by mr W last updated on 08/May/20

Commented by I want to learn more last updated on 08/May/20

$$\mathrm{Amin}$$

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