Question

help plz plz
don't spam ❌ ☹​

Answer 1

$\tt In\: \Delta QTR,\:\angle TRS \: is \: an \: exterior\:angle. \\ \\ \tt \therefore \angle QTR + \angle TQR = \angle TRS \\ \\ \tt \therefore \angle QTR = \angle TRS - \angle TQR \:\:\:\:\:\: ...(1) \\ \\ \tt For\: \Delta PQR , \: \angle QPR + \angle PQR = \angle PRS \\ \\ \tt as\:QT\:and\:RT\:are\:angle\:bisectors \\ \\ \tt \therefore \angle QPR +2 \angle TQR = 2 \angle TRS \\ \\ \tt \therefore \angle QPR = 2(\angle TRS - \angle TQR) \\ \\ \tt From\:equation\:1 \\ \\ \tt \therefore \angle QPR = 2 \angle QTR \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt \angle QTR = \frac{1}{2} \angle QPR }}}}$

प्रत्येक question ला sad emoji गरजेचं आहे का xD

Answer 2

pls talk with me param sundari )don't spam ❌ ☹