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प्रश्न
In the given figure, PB = PD. The value of x is ______.
विकल्प
85°
90°
25°
35°
उत्तर
In the given figure, PB = PD. The value of x is 25°.
Explanation:
∠PBE = ∠PDB + ∠BPD ......[Exterior angle property]
⇒ 120° = ∠PDB + θ ………….. (i)
Now, in ΔPBD, ∠PBD + ∠BPD + ∠PDB = 180° ......[Angle sum property]
⇒ ∠PBD + θ + ∠PDB = 180°
⇒ ∠PBD = 180° – 120° = 60° ......[Using (i)]
And PB = PD ......(Given)
∵ ∠PDB = ∠PBD = 60° ......(ii)
Now, in ΔPDC,
∠PDB = ∠DCP + ∠DPC ......[Exterior angle property]
⇒ 60° = x + 35° ......[Using (ii)] [∵ ∠DCP = x, ∠DPC = 35° (given)]
⇒ x = 60° – 35° = 25°
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