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Question

In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If AOQ = 58º, find ATQ. [CBSE 2015]

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Solution

It is given that AOQ = 58º.

We know that the angle subtended by an arc at the centre is twice the angle subtended by it any point on the remaining part of the circle.

ABQ=12AOQ=12×58°=29°

Now, AT is the tangent and OA is the radius of the circle through the point of contact A.

OAT=90° (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

In ∆ABT,

BAT+ABT+ATB=180° (Angle sum property)

90°+29°+ATB=180° BAT=OAT and ABT=ABQATB=180°-119°=61° ATQ=ATB=61°

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