The correct option is A 270∘
In the rhombus ABCD,
∠A+∠B+∠C+∠D=360∘ [Angle sum property]
In △AED and △AEB,AB=AD [Sides of rhombus]AE=AE [Common side]∠AED=∠AEB[Diagonals of a rhombus bisect at 90∘]∴△AED≅△AEB
∴∠DAC=∠CAB=12∠DAB
[CPCT]
⇒p=12∠A
⇒∠A=2p
Similarly,
∠B=2q, ∠C=2t and ∠D=2s
⇒2p+2q+2t+2s=360∘
⇒p+q+t+s=180∘
The diagonals of a rhombus intersect at 90∘
∴∠AEB=r=90∘
Therefore, p+q+r+s+t=180∘+90∘
⇒p+q+r+s+t=270∘