Given: Circle k(O), m NM =65° ∠PNO≅∠PMO Find: m∠PNO, m∠ONM

m∠PNO =16.25°  , m∠ONM=57.5°

Step-by-step explanation:

Given: Circle k(O), m NM =65° ∠PNO≅∠PMO

To find: m∠PNO, m∠ONM.

Since, central angle is equal to the measure of minor arc.

⇒∠MON = m NM =65°

In Δ MON , ON = OM [Radii of circle]

⇒ ∠ONM = ∠NMO   (i)   [Angle apposite to equal side of triangle are equal]

In Δ MON , ∠ONM + ∠NMO+∠MON =180°

⇒ ∠ONM + ∠ONM+65°=180°  [from (i)]

⇒ 2∠ONM=115°

⇒ ∠ONM=57.5°

⇒ ∠ONM = ∠NMO =57.5°

Also, an inscribed angle is half of a central angle that subtends the same arc.

⇒∠MPN =half of∠MON

=  

Also, ∠PNO≅∠PMO  [Given]

⇒∠PNO =∠PMO

⇒ ∠PNO +∠ONM =∠PMO+∠NMO   [∵∠ONM = ∠NMO]

⇒∠PNM=∠PMN

In ΔNPM

⇒ ∠MPN +∠PNM+∠PMN = 180°

⇒ 32.5° +∠PNM + ∠PNM= 180°

⇒ 2(∠PNM)= 147.5°

⇒ ∠PNM = 73.75°

Also,  ∠PNM = ∠PNO+∠ONM

⇒73.75°= ∠PNO+57.5°

⇒ ∠PNO =16.25°

Hence, m∠PNO =16.25°  , m∠ONM=57.5°

x=6 and x=3

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