Find the angles marked with a question mark shown in Fig. 17.27

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Solution

$In △ CEB : ∠ ECB + ∠ CBE + ∠ BEC = 180 ° (a n g l e s u m p r o p e r t y o f a t r i a n g l e) 40 ° + 90 ° + ∠ EBC = 180 ° ∴ ∠ EBC = 50 ° Also, ∠ EBC = ∠ ADC = 50 ° (opposite angle of a parallelogram) In △ FDC : ∠ FDC + ∠ DCF + ∠ CFD = 180 ° 50 ° + 90 ° + ∠ DCF = 180 ° ∴ ∠ DCF = 40 ° Now, ∠ BCE + ∠ ECF + ∠ FCD + ∠ F D C = 180 ° (in a parallelogram, the sum of alternate angles is 180 °) 50 ° + 40 ° + ∠ ECF + 40 ° = 180 ° ∠ ECF = 180 ° - 50 ° + 40 ° - 40 ° = 50 °$

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