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


Question

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

Options

A)
54o
B)
44o
C)
34o
D)
27o

The correct answer is A.

Explanation:

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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Dicussion (1)

  • 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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