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Question-94925




Question Number 94925 by I want to learn more last updated on 21/May/20
Commented by mr W last updated on 22/May/20
AB=AC=BC=Radius  ⇒ΔABC=equilateral  ⇒∠ABC=60°  similarly  ⇒∠ABD=60°  ⇒∠CBD=2×60=120°
$${AB}={AC}={BC}={Radius} \\ $$$$\Rightarrow\Delta{ABC}={equilateral} \\ $$$$\Rightarrow\angle{ABC}=\mathrm{60}° \\ $$$${similarly} \\ $$$$\Rightarrow\angle{ABD}=\mathrm{60}° \\ $$$$\Rightarrow\angle{CBD}=\mathrm{2}×\mathrm{60}=\mathrm{120}° \\ $$
Commented by I want to learn more last updated on 22/May/20
I appreciate sir
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$
Answered by ElOuafi last updated on 22/May/20
let α=∠CBD , β=∠BCD=∠BDC and   H the vertical dropping on CD  BC=BD⇒△BCD is isoscele in B     ⇒2β+α=π  in the triangle ABD   AD=AB ∧ BD=AB  ⇒△ABD is equilateral    ⇒∠ABD=∠BDA=∠DAB=(π/3)   ∴ in the triangle BHD ; (∠BHD=(π/2))      ⇒β+(π/3)+(π/2)=π  ⇒β=(π/6)     ⇒α=∠CBD=((2π)/3)
$${let}\:\alpha=\angle{CBD}\:,\:\beta=\angle{BCD}=\angle{BDC}\:{and}\: \\ $$$${H}\:{the}\:{vertical}\:{dropping}\:{on}\:{CD} \\ $$$${BC}={BD}\Rightarrow\bigtriangleup{BCD}\:{is}\:{isoscele}\:{in}\:{B} \\ $$$$\:\:\:\Rightarrow\mathrm{2}\beta+\alpha=\pi \\ $$$${in}\:{the}\:{triangle}\:{ABD} \\ $$$$\:{AD}={AB}\:\wedge\:{BD}={AB}\:\:\Rightarrow\bigtriangleup{ABD}\:{is}\:{equilateral} \\ $$$$\:\:\Rightarrow\angle{ABD}=\angle{BDA}=\angle{DAB}=\frac{\pi}{\mathrm{3}} \\ $$$$\:\therefore\:{in}\:{the}\:{triangle}\:{BHD}\:;\:\left(\angle{BHD}=\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\Rightarrow\beta+\frac{\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{2}}=\pi\:\:\Rightarrow\beta=\frac{\pi}{\mathrm{6}} \\ $$$$\:\:\:\Rightarrow\alpha=\angle{CBD}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by I want to learn more last updated on 22/May/20
I appreciate sir
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

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