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1.	भुजाओं 0? और २0 को क्रमश: बिन्दुओं 5 और पं=135° R ok £POT=110° &, /1 LPRQ T3-ASPA135°
Answer» Given,∠SPR =135° and ∠PQT =110° A.T.Q ∠SPR+∠QPR = 180° (by Linear pair axiom.) ⇒ 135°+∠QPR =180° ⇒ ∠QPR = 180° - 135° ⇒ ∠QPR = 45° ∠PQT+∠PQR = 180° (by Linear pair axiom.) ⇒ 110°+∠PQR =180° ⇒ ∠PQR = 70° Now,∠PQR+∠QPR+ ∠PRQ = 180° (Sum of the interior angles of thetriangle.)⇒70°+ 45° + ∠PRQ =180°⇒ 115°+ ∠PRQ =180° ⇒ ∠PRQ = 65° Like my answer if you find it useful! English mein kyo nahi pasand aa rahi

भुजाओं 0? और २0 को क्रमश: बिन्दुओं 5 और पं=135° R ok £POT=110° &, /1 LPRQ T3-ASPA135°

Answer»

Given,∠SPR =135° and ∠PQT =110°

A.T.Q

∠SPR+∠QPR = 180° (by Linear pair axiom.)

⇒ 135°+∠QPR =180°

⇒ ∠QPR = 180° - 135°

⇒ ∠QPR = 45°

∠PQT+∠PQR = 180° (by Linear pair axiom.)

⇒ 110°+∠PQR =180°

⇒ ∠PQR = 70°

Now,∠PQR+∠QPR+ ∠PRQ = 180° (Sum of the interior angles of thetriangle.)⇒70°+ 45° + ∠PRQ =180°⇒ 115°+ ∠PRQ =180°

⇒ ∠PRQ = 65°

Like my answer if you find it useful!

English mein kyo nahi pasand aa rahi