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Question-172483




Question Number 172483 by mnjuly1970 last updated on 27/Jun/22
Answered by mr W last updated on 28/Jun/22
Commented by mr W last updated on 28/Jun/22
set ∠BCE=α  ⇒EB^(⌢) =CA^(⌢)   CAE^(⌢) =CA^(⌢) +AE^(⌢) =EB^(⌢) +AE^(⌢) =AEB^(⌢)   ⇒CE=AB  ∠CED=90°  CE=CD cos (θ+α)  ⇒AB=CD cos (θ+α) ✓
$${set}\:\angle{BCE}=\alpha \\ $$$$\Rightarrow\overset{\frown} {{EB}}=\overset{\frown} {{CA}} \\ $$$$\overset{\frown} {{CAE}}=\overset{\frown} {{CA}}+\overset{\frown} {{AE}}=\overset{\frown} {{EB}}+\overset{\frown} {{AE}}=\overset{\frown} {{AEB}} \\ $$$$\Rightarrow{CE}={AB} \\ $$$$\angle{CED}=\mathrm{90}° \\ $$$${CE}={CD}\:\mathrm{cos}\:\left(\theta+\alpha\right) \\ $$$$\Rightarrow{AB}={CD}\:\mathrm{cos}\:\left(\theta+\alpha\right)\:\checkmark \\ $$
Commented by Tawa11 last updated on 28/Jun/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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