Find mPRˆ(arc). A. 115 B. 95 C. 125 D. 100

D. 100°

Step-by-step explanation:

Refer to picture attached

Lines QP and QR are tangent to radius of circle.

Triangles ΔQPO and ΔQRO are right triangles and line QO bisects angles of ∠PQR and ∠POR.

So as per sum of angles in right triangle, we have:

Then:

Correct answer choice is D. 100°

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.