Question

In the given figure, AB is the tangent of the circle
with centre O. If ∠DOA = 130°, then find the angle
∠DAB.

Answer 1

Answer:

Explanation:

Given:1. O is the centre of circle.

2.AB is a tangent from On point A of radius OA.

Angle DOA= 130

To Find: Angle DAB

Solution: In the circle,

O is the centre.

Therefore, OA and OD are radii of the circle.

Thus, OA=OD ....1

In triangle ODA,

OA =OD. (From 1)

Thus, triangle ODA is isoceles.

thus, angle ODA = angle OAD (isoceles triangle theorem)

thus let angle ODA = angle OAD = x ... 2

In triangle ODA,

angle DOA+ angle ODA+ angle OAD=180 (sum of the measures of the angles of a triangle is 180 degree)

angle DOA= 130,(Given)

thus ,

130+x+x=180

130+2x=180

2x=180-130

2x=50

x=25

Thus, angle OAD=25

AB is a tangent at point A on radius OA

Thus, angle OAB=90 (Tangent Theorem)

But angle OAB= angle OAD +angle DAB (angle addition property)

90= 25 +angle DAB

angle DAB= 90-25

angle DAB= 65

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