in fig, $o$ is the center of the circumference of $△ X Y Z$ . Tangents at $X$ and $Y$ intersects at $T$ . Given $∠ X T Y = 80^{0}$ and $∠ X O Z = 140^{0}$ . Calculate $∠ Z X Y$

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Solution

In $△ Z O X$
$\Rightarrow ∠ O X Z = ∠ O Z X$ $(O X = O Z = R a d i u s)$
$∠ Z X O = \frac{180 - 140}{2} = 20^{0}$
In Quadrilateral $T X O Y$
$\Rightarrow ∠ X O Y = 360 - (80 + 180)$
$= 100^{0}$
In $△ X O Y$
$\Rightarrow ∠ Y X O = ∠ O Y X$ $(O X = O Y = r a d i u s)$
$\Rightarrow ∠ Y X O = \frac{180 - 100}{2} = 40^{0}$
$\Rightarrow ∠ Z X Y = ∠ Z X O + ∠ Y X O$
$= 20 + 40$
$= 60^{0}$
Hence, the answer is $60^{0} .$

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