f.3 maths- quadrilaterals

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AB = AD and AB = AP , so AD = AP , so ㄥADP = ㄥAPD
In △ADP :

(ㄥADP + ㄥAPD) + (ㄥDAP) = 180

(ㄥADP x 2) + (ㄥDAB - ㄥPAB) = 180

ㄥADP x 2 + (90 - 60) = 180

ㄥADP = 75°

ㄥCPD + ㄥDPA + ㄥCPB + ㄥAPB = 360°

ㄥCPD + 75° + 75° + 60° = 360°

ㄥCPD = 150°


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By the first question , ㄥPSY = 75° ,

so ㄥYSX = ㄥPSX - 75° = 90 - 75 = 15°