Question

Ae khajur log
jaldi answer Karo .........​

Answer 1

$\huge\mathcal{Given}$

PQ is a chord of a circle with centre O.

$∠qpt = 60 {}^{0}$

let X be the point to the tangent PT.

$∠qpt + ∠opt = 90$

$↬∠opt = 30 {}^{0} - ∠qpt = 90 {}^{0} - 60 {}^{0} = 30 {}^{0}$

In ∆POQ

$∠poq = 180 - (∠opq - ∠pqo) = 180 - 30 - 30 = 120 {}^{0}$

Minor arc ∠POQ=120°

$∠poq = 360 {}^{0} - 120 {}^{0} = 240 {}^{0}$

Using tangent-secant theorem

$∠qrp =120 {}^{0}$

Answer 2

Answer:

Answer:

answer is 120

Step-by-step explanation:

given :PT is a tangent so ZOPT=90

ZQPT =60

ZOPQ = LOPT-ZQPT

ZOPQ=90-60

ZOPQ=30

ZOPQ+2OQP+POQ=D180

PROPERTY)

ZPOQ= 1803030

(ANGLE SUM

ZPOQ=120 Angle OPQ= angle OQP = 30° ie.,

Angle POQ= 120°. Also, angle PRQ= 12 reflex angle POQ.

ZPRQ=1/ 2(360-120)