Prove that tangent between two parallel …
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Sia ? 6 years, 6 months ago
Given, AB and CD are two parallel tangents to a circle with centre O.
From the figure we get,
AB⊥ST then {tex}\angle{/tex}ASQ = 90° and
CD⊥TS then {tex}\angle{/tex}CTQ = 90°
{tex}\angle{/tex}ASO = {tex}\angle{/tex}QSO = {tex}90^\circ \over 2{/tex}= 45°
Similarly, {tex}\angle{/tex}OTQ = 45°
Consider ΔSOT,
{tex}\angle{/tex}OTS = 45° and {tex}\angle{/tex}OST = 45°
{tex}\angle{/tex}SOT + {tex}\angle{/tex}OTS + {tex}\angle{/tex}OST = 180° (angle sum property)
{tex}\angle{/tex}SOT = 180° - ( {tex}\angle{/tex}OTS + {tex}\angle{/tex}OST) = 180° - (45° +45° )
= 180° - 90° = 90°
{tex}\therefore{/tex} {tex}\angle{/tex}SOT = 90o
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