Answer: 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
= [tex]\dfrac{65^{\circ}}{2}=32.5^{\circ}[/tex]
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°
