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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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