ISOSCELES TRIANGLE,Cosine rule in ΔBCD, –⇒ BC2 + DC2 = BD2 + 2 × BC × DC × cos120° [∵ all internal angles of HEXAGON is 120°]⇒ a2 + a2 = BD2 – 2 × a × a × cos60° [∵ cos(180° – 60°) = -cos 60°]⇒ BD2 = 3a2⇒ BD = √3aNow consider ΔABD,⇒ tan∠ADB = AB/BD⇒ tan∠ADB = a/√3a = 1/√3⇒ tan∠ADB = tan 30°∴ ∠ADB = 30°

"> ISOSCELES TRIANGLE,Cosine rule in ΔBCD, –⇒ BC2 + DC2 = BD2 + 2 × BC × DC × cos120° [∵ all internal angles of HEXAGON is 120°]⇒ a2 + a2 = BD2 – 2 × a × a × cos60° [∵ cos(180° – 60°) = -cos 60°]⇒ BD2 = 3a2⇒ BD = √3aNow consider ΔABD,⇒ tan∠ADB = AB/BD⇒ tan∠ADB = a/√3a = 1/√3⇒ tan∠ADB = tan 30°∴ ∠ADB = 30°

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If ABCDEF is a regular hexagon, then what is the value (in degrees) of ∠ADB?

SRMJEEE Analytical Geometry in SRMJEEE 1 year ago

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Consider ΔBCD is ISOSCELES TRIANGLE,Cosine rule in ΔBCD, –⇒ BC2 + DC2 = BD2 + 2 × BC × DC × cos120° [∵ all internal angles of HEXAGON is 120°]⇒ a2 + a2 = BD2 – 2 × a × a × cos60° [∵ cos(180° – 60°) = -cos 60°]⇒ BD2 = 3a2⇒ BD = √3aNow consider ΔABD,⇒ tan∠ADB = AB/BD⇒ tan∠ADB = a/√3a = 1/√3⇒ tan∠ADB = tan 30°∴ ∠ADB = 30°

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Posted on 03 Nov 2024, this text provides information on SRMJEEE related to Analytical Geometry in SRMJEEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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