Question

in a triangle PQR, PQ=5 cm, PR=7 cm, and QR=8 cm. if point S is on the circumscribed circle of the triangle such that PS bisects angle QPR, then the ratio of PS:RS is?​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$5\sqrt{7} /14$

Step-by-step explanation:

we know that

∵ PS is bisector

∴ PR/SR =PQ/ QS....eq1

and also given that QR=8

So, QS+SR=8....eq2

Solving eq1 and 2

SR= 14/3 and QS =10/3

length of angle bisector PS is $2\sqrt{ qrs(s-p)]}/(q+r)$  (given p=8,q=7,r=5)

s= (p+q+r)/2

 =(8+7+5)/2

  = 10

solving PS =$2\sqrt{ qrs(s-p)]}/(q+r)$

PS= 5√7/3

So PS :RS = $5\sqrt{7}/14$