in the adjoining figure, ABCD is a parallelogram. find the values of x, y and z
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Step-by-step explanation:
In tri.ACB,,
/_ACB = 50° --(alternate angle of /_CAD)
So,
/_CAB = "z" = 10° --(by property of triangle)
In tri.DAC,,
/_DCA =10° --(alternate angle of /_CAB)
Now,
in tri. CAD
/_C + /_ A + /_D = 180°
2x-2° + 50° + 10° =180°
2x -2 = 180-60
2x = 120+2
x = 122/2
x = 61°
/_y = /_CDA + /_DAC
/_y = 61° + 50°
/_y = 111°
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