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In the given figure, O is centre of circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠PQR. - Geometry Mathematics 2

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Question

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

Sum

Solution

Given, ∠QPR = 70°

m(arc PYR) = 160°

∠ = `1/2` m(arc PYR)

Now, m(arc QXR)

∠QPR = `1/2` m (arc QXR)

70° = `1/2` m(arc QXR)

∴ m(arc QXR) = 140°

Now, ∠QOR

using the properties of triangle, we have,

∠QOR = 2∠QPR

= 2 × 70°

= 140°

Now, ∠PQR = `1/2` m(arc PYR)   .....[Inscribed angle theorem]

∠PQR = `1/2 xx 160^circ`

∠PQR = 80°

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Inscribed Angle Theorem
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2018-2019 (March) Set 1

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


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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.


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

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

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

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

Find
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(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, ΔABC is an equilateral triangle. The angle bisector of ∠B will intersect the circumcircle ΔABC at point P. Then prove that: CQ = CA.


In the above figure, chord PQ and chord RS intersect each other at point T. If ∠STQ = 58° and ∠PSR = 24°, then complete the following activity to verify:

∠STQ = `1/2` [m(arc PR) + m(arc SQ)]

Activity: In ΔPTS,

∠SPQ = ∠STQ – `square`  ......[∵ Exterior angle theorem]

∴ ∠SPQ = 34°

∴ m(arc QS) = 2 × `square`° = 68°   ....... ∵ `square`

Similarly, m(arc PR) = 2∠PSR = `square`°

∴ `1/2` [m(arc QS) + m(arc PR)] = `1/2` × `square`° = 58°  ......(I)

But ∠STQ = 58°  .....(II) (given)

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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`

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