Respuesta :

cecepham

Answer:

a. 27

b. 153

c. 117

d. 207

e. 297

f.333

Step-by-step explanation:

In the circle, the center point is P, JM and NK are two diameters of the circle

=> JM intersects NK at point P

=> ∠MPN = ∠JPK

As P is on line segment NK

=> ∠MPN + ∠MPL + ∠LPK = 180°

Given that: ∠MPL = 63°; ∠LPK = 90°

=> ∠MPN + 63° + 90° = 180°

=> ∠MPN = 180° - 90° - 63° = 27°

=> ∠MPN = ∠JPK = 27°

As P is on line segment MK=J

=> ∠MPN + ∠NPJ = 180°

=> 27 + ∠NPJ = 180

=> m∠NPJ = 180 - 27 = 153°

As P is the center point of the circle and J, K, L, M, N are points on the circle, so that we have:

+) m arc JK = m∠JPK = 27°

+) m arc NJ = m∠NPJ = 153°

+) m arc JL = m∠JPL = m∠JPK + m∠KPL = 27 + 90 = 117°

+) m arc KNM = m arc KN + m arc NM = 180° + ∠MPN = 180 + 27 = 207°

+) m arc MJL = 360° - m arc ML = 360° - ∠MPL = 360 - 63 = 297°

+) m arc JLK = 360° - m arc JK = 360 - 27 = 333°

MrRoyal

The measures of the angles are:

  • Arc JK is 27 degrees.
  • Arc NJ is 153 degrees.
  • Arc JL is 117 degrees.
  • Angle KNM is 207 degrees.
  • Angle MJL is 297 degrees.
  • Angle JLK is 333 degrees.

From the attached image, we have:

  • The center of the circle is P.
  • The diameters are lines JM and NK.

The above highlights mean that:

[tex]\mathbf{\angle NPM \cong \angle JPK}[/tex]

[tex]\mathbf{\angle NPL = 90}[/tex]

So, we have:

[tex]\mathbf{\angle NPM +\angle MPL = 90}[/tex]

Substitute 63 for MPL

[tex]\mathbf{\angle NPM +63 = 90}[/tex]

Subtract 63 from both sides

[tex]\mathbf{\angle NPM = 27}[/tex]

[tex]\mathbf{\angle NPM \cong \angle JPK}[/tex] means that

[tex]\mathbf{\angle JPK = 27}[/tex]

So, arc JK is 27 degrees

Arc NJ is calculated using:

[tex]\mathbf{NJ = 180 - JK}[/tex] --- measure of angle in a semicircle

So, we have:

[tex]\mathbf{NJ = 180 - 27}[/tex]

[tex]\mathbf{NJ = 153}[/tex]

Arc JL is calculated using:

[tex]\mathbf{JL = 90+ JK}[/tex]

So, we have:

[tex]\mathbf{JL = 90 + 27}[/tex]

[tex]\mathbf{JL = 117}[/tex]

The measure of angle KNM is calculated using:

[tex]\mathbf{\angle KNM = 180 + \angle NPM}[/tex]

So, we have:

[tex]\mathbf{\angle KNM = 180 + 27}[/tex]

[tex]\mathbf{\angle KNM = 207}[/tex]

The measure of angle MJL is calculated using:

[tex]\mathbf{\angle MJL = 180 + JK + 90}[/tex]

So, we have:

[tex]\mathbf{\angle MJL = 180 + 27 + 90}[/tex]

[tex]\mathbf{\angle MJL = 297}[/tex]

Lastly, the measure of angle JLK is calculated using:

[tex]\mathbf{\angle JLK = 180 + \angle MPL + 90}[/tex]

So, we have:

[tex]\mathbf{\angle JLK = 180 + 63 + 90}[/tex]

[tex]\mathbf{\angle JLK = 333}[/tex]

Read more about circles, central angles, arcs, and chord at:

https://brainly.com/question/3670983

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