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Question-179927




Question Number 179927 by Tawa11 last updated on 04/Nov/22
Commented by som(math1967) last updated on 04/Nov/22
Commented by som(math1967) last updated on 04/Nov/22
 3b  ∠ABX=∠AEX=((180−108)/2)=36  same way ∠ADE=36  ∠DEX=108−36=72  ∴∠AXB=∠DXE=180−36−72=72  ∠BAX=108−36=72  ∴ angles of △ABX are  36,72°,72°
$$\:\mathrm{3}\boldsymbol{{b}} \\ $$$$\angle{ABX}=\angle{AEX}=\frac{\mathrm{180}−\mathrm{108}}{\mathrm{2}}=\mathrm{36} \\ $$$${same}\:{way}\:\angle{ADE}=\mathrm{36} \\ $$$$\angle{DEX}=\mathrm{108}−\mathrm{36}=\mathrm{72} \\ $$$$\therefore\angle{AXB}=\angle{DXE}=\mathrm{180}−\mathrm{36}−\mathrm{72}=\mathrm{72} \\ $$$$\angle{BAX}=\mathrm{108}−\mathrm{36}=\mathrm{72} \\ $$$$\therefore\:{angles}\:{of}\:\bigtriangleup{ABX}\:{are} \\ $$$$\mathrm{36},\mathrm{72}°,\mathrm{72}° \\ $$
Commented by Tawa11 last updated on 04/Nov/22
God bless you sir. I appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by som(math1967) last updated on 04/Nov/22
   For R48  In △ADB and △BDC  ∠ADB=∠BDC  ∠CDB=∠CBD  ∴ ∠A=∠C  again ABCD  is cyclic  ∴∠A+∠C=180  ⇒2∠A=180  ∴∠A=90  ∴ BD is diameter
$$\:\:\:{For}\:{R}\mathrm{48} \\ $$$${In}\:\bigtriangleup{ADB}\:{and}\:\bigtriangleup{BDC} \\ $$$$\angle{ADB}=\angle{BDC} \\ $$$$\angle{CDB}=\angle{CBD} \\ $$$$\therefore\:\angle{A}=\angle{C} \\ $$$${again}\:{ABCD}\:\:{is}\:{cyclic} \\ $$$$\therefore\angle{A}+\angle{C}=\mathrm{180} \\ $$$$\Rightarrow\mathrm{2}\angle{A}=\mathrm{180} \\ $$$$\therefore\angle{A}=\mathrm{90} \\ $$$$\therefore\:{BD}\:{is}\:{diameter} \\ $$
Commented by Tawa11 last updated on 04/Nov/22
God bless you sir. I appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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