Question

please find this answer

Answer 1

use angle sum property of trianglePRQ
the use linear pair<trs+<prq =180 then solve

Answer 2

(1)
In ∆PQR, sum of all ∠s = 180°

=>∠PQR + ∠QPR + ∠PRQ = 180...(angle sum property)

=> 15 + 35 + ∠PRQ = 180
=> 50 + ∠PRQ = 180
=> ∠PRQ = 180 - 50
=> ∠PRQ = 130

Hence, ∠PRQ is 130°.....(equation 1)

(2)
Now, QS is a straight line and ray PR stands on it.
Therefore, ∠PRQ + ∠PRS = 180°...(linear pair property)
From eq 1,

=> 130 + ∠PRS = 180
=> ∠PRS = 180 - 130
=> ∠PRS = 50°

Since T is a part of line PR, hence ∠TRS is 50°.....(equation 2)

(3)
We know that in any ∆, the exterior angle formed by producing one side is equal to the sum of opposite interior angles.
Hence,
In ∆ TRS,
∠PTS = ∠TSR + ∠TRS
From equation 2,

=> ∠PTS = 50 + 50
=> ∠PTS = 100

Hence, ∠PTS is 100°....(equation 3)

From equations 1, 2 and 3, you get your answer.

Hope it'll help.. :-D