Question

In the circle with centre O, PT is a tangent at P. PQ is a chord and if angle TPQ=x degree. find angle POQ in terms of x.

Answer 1

Answer:

Your answer is:

Since OP and OQ are the radii

therefore both are equal.

PT is parallel to OQ so OP formed between two parallel lines represent height

height is 90°

so Angle OPQ and OQP are 45° ....[Isosceles triangle theorem]

therefore angle TPQ is 45°

[Alternate angles property]

plz mark it as brainliest

Answer 2

given,

angle TPQ=x

to find,

angle POQ

proof,

in triangle POQ

OP=OQ(Radius of circle)

=>POQ is an isoceless triangle

therefore angle OPQ=PQO

now,

angle OPT=90°(Radius intersecting the tangent at point of contact is perpendicular to the tangent)

also, Angle OPT=OPQ+QPT...... (i)

substituting values of OPT, QPT in (i)

90°=OPQ+x

=OPQ=90°-x

now in triangle POQ

POQ+OPQ+PQO=180°( ANGLE SUM PROPERTY)

Substituting OPQ IN ABOVE

POQ+(90°-x)+(90°-x)=180°

POQ=180°-(90°-x)-(90°-x)

POQ=180°-90°-90°+x+x

POQ=2x

HOPE IT HELPS YOU MATE!