In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.
54o
44o
34o
27o
Correct answer is A
In the diagram above, x1 = 180o - 117o = 63o(opposite angles of a cyclic quad.)
x1 = x2 (base angles of isos. \(\Delta\))
x1 + x2 + \(\alpha\) = 180o (sum of angles of a \(\Delta\)
63o + 63o + \(\alpha\) = 180o
\(\alpha\) = 180o - (63 + 63)o
= 54o
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