Maths MCQ - Quadrilaterals

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 = 180⁰ (As we know each angle of regular pentagon is 180⁰.)
Now, from angle sum property in △ABE,
∠BEA + ∠EBA + ∠BAE = 180⁰,
So, ∠BEA + ∠BEA + 180⁰ = 180⁰,
2∠BEA = 72⁰
∠BEA = 36⁰
∠BEA = ∠EBA 36⁰ ....... (i)
Similarly, in △DEC,
∠CED = ∠ECD 36⁰ ....... (ii)
And in △CDB,
∠BDC = ∠DBC = 36⁰ ....... (iii)
Then,
∠DEA = ∠BEA + ∠CEB + ∠CED
We know, ∠DEA = 180⁰(Angle of regular pentagon)
Then, 36⁰ + ∠CEB + 36⁰ = 108⁰
∠CEB = 36⁰
Similarly, ∠CEB = ∠CAD
= ∠ADB = ∠ACE
= ∠DBE = 36⁰
And sum of all 5 vertical angles of star = 36⁰ + 36⁰ + 36⁰ + 36⁰ + 36⁰ = 180⁰

From the figure, AD ∥ BC

So, ∠ADC = ∠BCR = 120⁰ [corresponding angles]
∴ ∠BCS = 180⁰ - 120⁰ = 60⁰
Similarly, SP ∥ RQ
∠RQP = ∠SPB = 70⁰ [corresponding angles]
Now, DC ∥ AP and SP is a transversal
∴ ∠RSP = 70⁰ [alternate angles]
In △OSC,
∠OSC + ∠SCO + ∠COS = 180⁰
⇒ 70⁰ + 60⁰ + ∠COS = 180⁰
⇒ ∠COS = θ = 50⁰