English

In the figure, a circle with center C has m(arc AXB) = 100° then find central ∠ACB and measure m(arc AYB). - Geometry Mathematics 2

Advertisements
Advertisements

Question

In the figure, a circle with center C has m(arc AXB) = 100° then find central ∠ACB and measure m(arc AYB).

Sum

Solution


In the given figure, m(arc AXB) = ∠ACB  ......[Definition of measure of minor arc]

∴ ∠ACB = 100° 

m(arc AXB) + m(arc AYB) = 360°   ......[Measure of complete circle is 360°]

∴ 100° + m(arc AYB) = 360°

∴ m(arc AYB) = 360° – 100°

∴ m(arc AYB) = 260°

shaalaa.com
Inscribed Angle Theorem
  Is there an error in this question or solution?
Chapter 3: Circle - Q.4

RELATED QUESTIONS

In the adjoining figure chord EF || chord GH.
Prove that chord EG ≅ chord FH.
Fill in the boxes and write the complete proof.


In the given figure, O is the centre of the circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following m(arc QXR).


In the given figure, O is centre of circle. ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠QOR.


In the given figure, O is centre of circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠PQR.


A circle with centre P is inscribed in the ABC. Side AB, side BC and side AC touch the circle at points L, M and N respectively.  Radius of the circle is r. 

Prove that: `"A" (triangle "ABC") =1/2 ("AB" + "BC" + "AC") xx "r"`


In the figure, if the chord PQ and chord RS intersect at point T, prove that: m∠STQ = `1/2` [m(arc PR) + m(arc SQ)] for any measure of ∠STQ by filling out the boxes

Proof: m∠STQ = m∠SPQ + `square`  .....[Theorem of the external angle of a triangle]

 = `1/2` m(arc SQ) + `square`   .....[Inscribed angle theorem]

= `1/2 [square + square]`


Prove that angles inscribed in the same arc are congruent.


Given: In a circle with center C, ∠PQR and ∠PSR are inscribed in same arc PQR. Arc PTR is intercepted by the angles.

To prove: ∠PQR ≅ ∠PSR.

Proof:

m∠PQR = `1/2 xx ["m"("arc PTR")]`         ......(i) `square`

m∠`square` = `1/2 xx ["m"("arc PTR")]`   ......(ii) `square`

m∠`square` = m∠PSR      .....[By (i) and (ii)]

∴ ∠PQR ≅ ∠PSR


In the figure, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.


Proof:

Draw seg OD.

∠ACB = `square`     ......[Angle inscribed in semicircle]

∠DCB = `square`    ......[CD is the bisector of ∠C]

m(arc DB) = `square`   ......[Inscribed angle theorem]

∠DOB = `square`   ......[Definition of measure of an arc](i)

seg OA ≅ seg OB     ...... `square` (ii)

∴ Line OD is `square` of seg AB    ......[From (i) and (ii)]

∴ seg AD ≅ seg BD


In the figure, chord LM ≅ chord LN, ∠L = 35°. 

Find
(i) m(arc MN)

(ii) m(arc LN)


In the figure, if O is the center of the circle and two chords of the circle EF and GH are parallel to each other. Show that ∠EOG ≅ ∠FOH


In the figure, the centre of the circle is O and ∠STP = 40°.

  1. m (arc SP) = ? By which theorem?
  2. m ∠SOP = ? Give reason.


In the above figure, ∠L = 35°, find :

  1. m(arc MN)
  2. m(arc MLN)

Solution :

  1. ∠L = `1/2` m(arc MN) ............(By inscribed angle theorem)
    ∴ `square = 1/2` m(arc MN)
    ∴ 2 × 35 = m(arc MN)
    ∴ m(arc MN) = `square`
  2. m(arc MLN) = `square` – m(arc MN) ...........[Definition of measure of arc]
    = 360° – 70°
    ∴ m(arc MLN) = `square`

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×