The sum of the interior angles of a pentagon is:
(5−2)×180=540∘(5 - 2)\times 180 = 540^\circ(5−2)×180=540∘
Given that ∠A=∠B=90∘\angle A = \angle B = 90^\circ∠A=∠B=90∘, we have:
A+B+C+D+E=540∘A + B + C + D + E = 540^\circA+B+C+D+E=540∘ 90+90+C+D+E=540⇒C+D+E=360∘90 + 90 + C + D + E = 540 \Rightarrow C + D + E = 360^\circ90+90+C+D+E=540⇒C+D+E=360∘
Since the pentagon is equilateral and symmetric (with AAA and BBB equal and adjacent), we can infer:
C=EC = EC=E
Let C=E=xC = E = xC=E=x and D=yD = yD=y. Then:
2x+y=3602x + y = 3602x+y=360
Because the pentagon is concave, one interior angle must be greater than 180∘180^\circ180∘, which is DDD. From geometric reasoning, the valid solution is:
x=135∘,y=90∘x = 135^\circ,\quad y = 90^\circx=135∘,y=90∘
Therefore:
∠E=135∘\angle E = 135^\circ∠E=135∘
Answer: 135∘135^\circ135∘
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