Question

plz.. solve I Will Mark it as brainlist...

Answer 1

mark me as a brainlist

Answer 2

Answer:

Join OP and OQ .

In  triangle OPS and triangle OQS : -

OP= OQ……Radius of circle

OPS= OQS= 90 degree (Tangent is to the radius through the point of contact)

OS=OS….(common )

Therefore, by Right angle-Hypotenuse-Side criterion of congruence, we have

triangle OPS is congruent to triangle  OQS ( by RHS)

The corresponding parts of the congruent triangles are congruent.

angle POS = angle QOS……cpct,

angle OSP = angle  OSQ ……..cpct

In quadrilateral PSQO:

angle POQ+ angle PSQ=180 degree ( as the total sum of interior angles in a quadrilateral=360 degree )

angle POQ = angle PSQ=900 degree

So  OSP = OSQ = 45 degree, As  angle PSQ=900 degree

STC=90 degree ( sum of two co interior angles is 180 degree   )

SOT is an isosceles triangle having two sides:

OS=OT

Similarly, QOT = 45 degree

SOT = 90 degree

hence proved

hey...i have done a great hard work for this answer pllzz appreciate me and make it as brainliest

Step-by-step explanation: