Question

in the diagram, TA is a tangent to the circle at A. If BCA=56° and DAT=60°, find BA^D

Answer 1

Answer:

In order to prove that △ADT is isosceles i.e., TA = TD, it is sufficient to show that ∠TAD = ∠TDA.

Since ∠TAB and ∠BCA are angles in the alternate segments of chord AB.

∴ ∠TAB = ∠BCA …….(i)

It is given that AD is the bisector of ∠BAC.

∴ ∠BAD = ∠CAD ……..(ii)

Now, ∠TAD = ∠TAB + ∠BAD

⇒ ∠TAD = ∠BCA + ∠CAD [Using (i) and (ii)]

⇒ ∠TAD = ∠DCA + ∠CAD [∵ ∠BCA = ∠DCA]

⇒ ∠TAD = 180˚ - ∠CDA [In △CAD, ∠CAD + ∠DCA + ∠CDA = 180˚]

∴ ∠CAD + ∠BCA = 180˚ - ∠CDA]

⇒ ∠TAD = ∠TDA [∵ ∠CDA + ∠TDA = 180˚]

⇒ TD = TA

Hence, △ADT is isosceles .

Hope it helps you ....

Answer 2

Angle BAD is equal to 64°.

Step-by-step explanation:

• When a line touches at a point on the circumference of the circle and is perpendicular to the radius of the circle, the line is known as tangent of the circle and the point of tangency is the meeting point of the tangent at the circumference of the circle.
• Here, according to the given information,  TA is a tangent to the circle at A and angles BCA=56° and DAT=60°.
• Then angle ACD= 60° as angle ACD is an alternate angle.
• Also, angle BCD = angle BCA + angle ACD, that is, angle BCD = 56 + 60 = 116°.
• Now, angle BAD = 180 - 116 = 64°, since they are angles in a cyclic quadrilateral.

Hence, angle BAD is equal to 64°.

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