# For ΔPQR and ΔXYZ, the correspondence PQR ↔ YZX is a similarity. m∠P = 2m∠Q and m∠X = 120. Find m∠Y.

Given, ΔPQR and ΔXYZ

P = 2Q and X = 120

If the correspondence PQRYZX is a similarity then

mP = mX ... (i)

mQ = mY ... (ii)

mR = mZ ... (iii)

From eq. (i) and (ii)

P = 120

Because mP = 2mQ

Hence, 2Q = 120

Q = 60

And Q = Y

Hence, Y = 60.

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