Question

refer to attachment​

Answer 1

Answer:

Step-by-step explanation:

Proof:

∠PQR +∠PQS =180° (by Linear

Pair axiom)

∠PQS

=180°– ∠PQR

— (i)

∠PRQ

+∠PRT

= 180° (by Linear Pair axiom)

∠PRT

= 180° – ∠PRQ

 ∠PRQ=180°– ∠PQR — (ii)

 [∠PQR = ∠PRQ]

From (i) and (ii)

∠PQS

= ∠PRT

= 180°– ∠PQR

∠PQS

= ∠PRT

 

Hence, ∠PQS = ∠PRT

Hope it helps you

Answer 2

FIGURE in the attachment

GIVEN :-

$\sf \angle PQR = \angle PRQ$

➡ ∠1 = ∠2 -------(i)

to prove :-

$\sf \angle PQS = \angle PRT$

proof :-

» ∠1 + ∠3 = 180° (linear pair)

➡ ∠3 = 180° - ∠1 ------(ii)

similarly,

» ∠2 + ∠4 = 180° (linear pair)

➡ ∠4 = 180° - ∠2

from equation (i), we get

➡ ∠4 = 180° - ∠1 (since ∠1 = ∠2) ------(iii)

from equation (ii) and (iii)

➡ ∠3 = ∠4

➡ ∠PQS = ∠PRT

hence proved!