In the diagram, < QPR = 60o
< PQR = 50...
< PQR = 50...
In the diagram, < QPR = 60o
< PQR = 50o
< QRS = 2xo
< SRP = 3xo
< UQP = yo and RS//TU
calculate y
A.
102o
B.
78o
C.
70o
D.
60o
Correct answer is A
< RQT = < QRS = 2x (Alternate angles). But in \(\bigtriangleup\) PRQ
50 + 60 + (2x + 3x) = 180o(Angles in a triangle)
110 + 5x = 180o
5x = 180o - 110 = 70o
x = \(\frac{70}{5}\) = 14o
Also y + 50 + 2(14) = 180o
y + 50 + 28 = 180o
y = 180 - 78
y = 102o
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