Question

. In the given figure, AD is a diameter of a circle with centre O and AB is a tangent at A. C is a point on the circle such that DC produced intersects the tangent at B and ∠ABC = 50o . Find ∠AOC.

Answer 1

Step-by-step explanation:

angle abc= 50°

angle bad=90° because radius to tangent is perpendicular

angle bda=40°asp

angle aoc=80° because of degree measure theorem

Answer 2

$\angle AOC=80^{\circ}$

Step-by-step explanation:

$\angle ABC=50^{\circ}$

AD is a diameter of circle.

We know that radius is perpendicular to tangent.

Therefore, OA is perpendicular to AB.

$\angle OAC=90^{\circ}$

In triangle ABD,

$\angle ADB+\angle DAB+\angle DBA=180^{\circ}$

Substitute the values then we get

$\angle ADB+90+50=180$

$\angle ADB+140=180$

$\angle ADB=180-140=40$ degrees

We know that central angle is twice the inscribed angle

$\angle AOC=2\times \angle ADC=2\times 40=80^{\circ}$

Hence,$\angle AOC=80^{\circ}$

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