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Question-165570




Question Number 165570 by cortano1 last updated on 04/Feb/22
Answered by Rohit143Jo last updated on 04/Feb/22
θ=15°
$$\theta=\mathrm{15}° \\ $$
Answered by Rohit143Jo last updated on 04/Feb/22
ABCD is a square, AB=BC=CD=DA                                   and, each angle 90°        Now, In ΔCDE, DE=CE=CD                                     So, ∠DCE=∠CDE=∠CED=60°                   ∴ΔCDE is equilateral triangle.        Now, In ΔBCE, BC=CE                                     So, ∠BEC=∠CBE=x (let)                              ∴ x+x+30°=180°                                ⇒ x=75°        Now, ∠ABC=90°                ⇒ θ+75°=90°                ⇒ θ=15°
$${ABCD}\:{is}\:{a}\:{square},\:{AB}={BC}={CD}={DA}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and},\:{each}\:{angle}\:\mathrm{90}° \\ $$$$\:\:\:\:\:\:{Now},\:{In}\:\Delta{CDE},\:{DE}={CE}={CD} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{So},\:\angle{DCE}=\angle{CDE}=\angle{CED}=\mathrm{60}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\Delta{CDE}\:{is}\:{equilateral}\:{triangle}. \\ $$$$\:\:\:\:\:\:{Now},\:{In}\:\Delta{BCE},\:{BC}={CE} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{So},\:\angle{BEC}=\angle{CBE}={x}\:\left({let}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:{x}+{x}+\mathrm{30}°=\mathrm{180}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{x}=\mathrm{75}° \\ $$$$\:\:\:\:\:\:{Now},\:\angle{ABC}=\mathrm{90}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\theta+\mathrm{75}°=\mathrm{90}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\theta=\mathrm{15}° \\ $$
Commented by Tawa11 last updated on 05/Feb/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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