In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177^o. Calculate < PST.

54^o

44^o

34^o

27^o

Correct answer is A

In the diagram above, x₁ = 180^o - 117^o = 63^o(opposite angles of a cyclic quad.)

x₁ = x₂ (base angles of isos. \(\Delta\))

x₁ + x₂ + \(\alpha\) = 180^o (sum of angles of a \(\Delta\)

63^o + 63^o + \(\alpha\) = 180^o

\(\alpha\) = 180^o - (63 + 63)^o

= 54^o

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In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.

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In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177^o. Calculate < PST.