Q11 of 47 Page 10

In Fig., AB is a diameter of a circle with centre O and AT is a tangent. If ∠AOQ=58°, find ∠ATQ.

Given: AB is a diameter


∠AOQ=58°


To find: The value of ∠ATQ.


Theorem Used:


Tangent to a circle at a point is perpendicular to the radius through the point of contact.


Explanation:



As we know angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point of the remaining part.


∴ ∠AOQ = 2∠ABQ


⇒ ∠ABQ = 1/2 ∠AOQ


⇒ ∠ABQ = 1/2 58°


⇒ ∠ABQ = 29°


⇒ ∠ABT = 29°


As AT is the tangent to radius OA,


By the theorem stated OA ⊥ AT.


∴ ∠OAT = 90°


⇒ ∠BAT = 90°


In Δ BAT


∠BAT + ∠ABT + ∠ATB = 180°


⇒ 90° + 29° + ∠ATB = 180°


⇒ 119° + ∠ATB = 180°


⇒ ∠ATB = 61°


So ∠ATQ = 61°


More from this chapter

All 47 →
9

Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle.

 10

Prove that a parallelogram circumscribing a circle is a rhombus.

 12

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

 13

AB is a diameter and AC is a chord of a circle with center O such that BAC = 30°. The tangent at C intersects extended AB at a point D. Prove that BC = BD.