165k views
2 votes
Given: Circle k(O), m NM =65° ∠PNO≅∠PMO Find: m∠PNO, m∠ONM

Given: Circle k(O), m NM =65° ∠PNO≅∠PMO Find: m∠PNO, m∠ONM-example-1
User Paolo M
by
7.1k points

2 Answers

2 votes

Answer:

Answer: m∠PNO =16.25° , m∠ONM=57.5°

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°

Explanation:

User Lkkkk
by
7.3k points
2 votes

Answer: m∠PNO =16.25° , m∠ONM=57.5°

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

=
(65^(\circ))/(2)=32.5^(\circ)

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°

User Austin Crim
by
6.5k points