Circle

In the figure, $PT$ is a tangent to the circle at point $T$. $CT$ is a diameter.

$\angle ABC = 70^\circ$, $\;$ $\angle ACB = 50^\circ$.

Calculate the measure of
  1. $\angle CBT$
  2. $\angle BAT$
  3. $\angle APT$


$\angle ABC = 70^\circ$, $\;$ $\angle ACB = 50^\circ$

$PT =$ tangent to the circle at point $T$

$CT =$ diameter of the circle

  1. $\angle CBT = 90^\circ$ $\;\;\;$ [angle in a semicircle]


  2. In $\;$ $\triangle ABC$, $\;$ $\angle CAB + \angle ABC + \angle ACB = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

    $\therefore \;$ $\angle CAB = 180^\circ - \left(\angle ABC + \angle ACB\right) = 180^\circ - \left(70^\circ + 50^\circ\right) = 60^\circ$

    $\angle CAT = 90^\circ$ $\;\;\;$ [angle in a semicircle]

    Now, $\;$ $\angle CAT = \angle CAB + \angle BAT$

    $\therefore \;$ $\angle BAT = \angle CAT - \angle CAB = 90^\circ - 60^\circ = 30^\circ$


  3. $\angle CTB = \angle CAB = 60^\circ$ $\;\;\;$ [angles of the same segment]

    In the figure, $\;$ $\angle CBT = \angle ABC + \angle PBT$

    $\therefore \;$ $\angle PBT = \angle CBT - \angle ABC = 90^\circ - 70^\circ = 20^\circ$

    Let $PB$ and $CT$ intersect at point $X$.

    $\because \;$ $P - X - B$ are collinear points, $\;$ $\angle XBT = \angle PBT = 20^\circ$

    $\because \;$ $C - X - T$ are collinear points, $\;$ $\angle XTB = \angle CTB = 60^\circ$

    In $\;$ $\triangle XTB$, $\;$ $\angle XTB + \angle XBT + \angle TXB = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

    $\therefore \;$ $\angle TXB = 180^\circ - \left(\angle XTB + \angle XBT\right) = 180^\circ - \left(60^\circ + 20^\circ\right) = 100^\circ$

    Now, $\;$ $\angle TXB + \angle CXB = 180^\circ$ $\;\;\;$ [linear pair]

    $\therefore \;$ $\angle CXB = 180^\circ - \angle TXB = 180^\circ - 100^\circ = 80^\circ$

    $\angle PXT = \angle CXB = 80^\circ$ $\;\;\;$ [vertically opposite angles]

    $\angle PTX = 90^\circ$ $\;\;\;$ [angle between diameter and tangent is $90^\circ$]

    In $\triangle PXT$, $\;$ $\angle PXT + \angle PTX + \angle XPT = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

    $\therefore \;$ $\angle XPT = 180^\circ - \left(\angle PXT + \angle PTX\right) = 180^\circ - \left(80^\circ + 90^\circ\right) = 10^\circ$

    $\because \;$ $P - A - X$ are collinear points, $\;$ $\angle APT = \angle XPT = 10^\circ$

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