In the figure (not drawn to scale) given below, if AD = CD = BC and BCE ∠75°, how much is the value of ∠DBC?
Explanation:
In ∆ADC, Since AD = CD ⇒ ∠CAD = ∠ACD = x Also, ∠CDB = 2x (Exterior angle theorem)
Now, ∠ACD + ∠DCB + ∠BCE = 180° ⇒ x + y + 75 = 180 ⇒ x + y = 105 ….(1)
In ∆DCB, Since, CD = CB ⇒ ∠DBC = 2x
∴ 2x + 2x + y = 180 ⇒ 4x + y = 180 …(2)
Solving (1) and (2) we get x = 25° and y = 80°
⇒ ∠DBC = 2x = 50°
Hence, option (c).
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