- Option : B
- Explanation : (B) From the given information, we can form the figure as,

Now, in △ABE, AB = AE (As these are the sides of regular pentagon)

So, from base angle theorem, we know

∠BEA = ∠EBA

∠BAE = 180^{0}(As we know each angle of regular pentagon is 180^{0}.)

Now, from angle sum property in △ABE,

∠BEA + ∠EBA + ∠BAE = 180^{0},

So, ∠BEA + ∠BEA + 180^{0}= 180^{0},

2∠BEA = 72^{0}

∠BEA = 36^{0}

∠BEA = ∠EBA 36^{0}....... (i)

Similarly, in △DEC,

∠CED = ∠ECD 36^{0}....... (ii)

And in △CDB,

∠BDC = ∠DBC = 36^{0}....... (iii)

Then,

∠DEA = ∠BEA + ∠CEB + ∠CED

We know, ∠DEA = 180^{0}(Angle of regular pentagon)

Then, 36^{0}+ ∠CEB + 36^{0}= 108^{0}

∠CEB = 36^{0}

Similarly, ∠CEB = ∠CAD

= ∠ADB = ∠ACE

= ∠DBE = 36^{0}

And sum of all 5 vertical angles of star = 36^{0}+ 36^{0}+ 36^{0}+ 36^{0}+ 36^{0}= 180^{0}

2. In the given figure, both ABCD and PQRS are the parallelograms. The value of θ will be

- Option : A
- Explanation : (A) From the figure, AD ∥ BC

So, ∠ADC = ∠BCR = 120^{0}[corresponding angles]

∴ ∠BCS = 180^{0}- 120^{0}= 60^{0}

Similarly, SP ∥ RQ

∠RQP = ∠SPB = 70^{0}[corresponding angles]

Now, DC ∥ AP and SP is a transversal

∴ ∠RSP = 70^{0}[alternate angles]

In △OSC,

∠OSC + ∠SCO + ∠COS = 180^{0}

⇒ 70^{0}+ 60^{0}+ ∠COS = 180^{0}

⇒ ∠COS = θ = 50^{0}

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