Q5) O is the circumference of the triangle ABC and Odis perpendicular on BC. Prove that ∠BOD = ∠A.
Solution:
Given O is the circum centre of triangle ABC and OD⊥BC
to prove ∠BOD = 2 ∠A
Pro of :
In ΔOBD and ΔOCD
∠ODB = ∠ODC [Each900]
OB = OC [Radius of circle]
OD = OD [Common]
ThenΔOBD≅ΔOCD [ by RHS Condition] .
∴ ∠BOD = ∠COD…..(i)
[PCT] . by degree measure theorem ∠BOC = 2 ∠BAC
⇒ 2 ∠BOD = 2 ∠BAC [ by using(i)]
⇒ ∠BOD = ∠BAC.
Q6) In figure 16.135, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.
Solution:
Given, BO is the bisec to r of ∠ABC
to prove AB = BC
Pro of :
Since, BO is the bisec to r of ∠ABC.
Then, ∠ABO = ∠CBO…..(i)
Since, OB = OA [Radius of circle]
Then, ∠ABO = ∠DAB…..(ii) [opposite Angles to equal sides]
Since OB = OC [Radius of circle]
Then, ∠OAB = ∠OCB…..(iii) [opposite Angles to equal sides]
Compare equations (i), (ii) and (iii)
∠OAB = ∠OCB…..(iv)
InΔOAB andΔOCB
∠OAB = ∠OCB [From (iv)]
∠OBA = ∠OBC [Given]
OB = OB [Common]
Then, ΔOAB≅ΔOCB [ by AAScondition]
∴AB = BC [CPCT]
Q7) In figure 16.136, O is the centre of the circle, then prove that ∠x = ∠y+ ∠z.
Solution:
We have,
∠3 = ∠4 [ Angles insamesegment]
∴ ∠x = 2 ∠3 [ by degree measure theorem]
⇒ ∠x = ∠3+ ∠3
⇒ ∠x = ∠3+ ∠4…..(i) [ ∠3 = angle4]
But ∠y = ∠3+ ∠1 [ by exteriorangleproperty]
⇒ ∠3 = ∠y− ∠1…..(ii)
from(i) and(ii)
∠x = ∠y− ∠1+ ∠4
⇒ ∠x = ∠y+ ∠4− ∠1
⇒ ∠x = ∠y+ ∠z+ ∠1− ∠1 [ by exteriorangleproperty]
⇒ ∠x = ∠y+ ∠z
Q8) In figure 16.137, O and O’ are centers of two circles intersecting at B and C. ACD is a straight line, find x.
Solution:
by degree measure theorem ∠AOB = 2 ∠ACB
⇒ 130^{0} = 2 ∠ACB
⇒ ∠ACB = = 65^{0}
∴ ∠ACB+ ∠BCD = 180^{0} [Linerapair of Angles ]
⇒ 65^{0}+ ∠BCD = 180^{0}
⇒ ∠BCD = 180^{0}−65^{0 }= 115^{0} by degree measure theoremreflex ∠BOD = 2 ∠BCD
⇒ reflex ∠BOD = 2×115^{0} = 230^{0}
Now,reflex ∠BOD+ ∠BO′D = 360^{0} [Complexangle]
⇒ 230^{0}+x = 360^{0}
⇒ x = 360^{0}−230^{0}
∴x = 130^{0}
Q9) In figure 16.138, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40^{0},find ∠RTS..
Solution:
Since PQ is diameter
Then,
∠PRQ = 90^{0} [Angleinsemicircle]
∴ ∠PRQ+ ∠TRQ = 180^{0} [Linearpair of angle]
90^{0}+ ∠TRQ = 180^{0}
∠TRQ = 180^{0}−90^{0} = 90^{0}.
by degree measure theorem
∠ROS = 2 ∠RQS
⇒ 40^{0} = 2 ∠RQS
⇒ ∠RQS = = 20^{0}
In ΔRQT, by Angle sum property
∠RQT+ ∠QRT+ ∠RTS = 180^{0}
⇒ 20^{0}+90^{0}+ ∠RTS = 180^{0}
⇒ ∠RTS = 180^{0}−20^{0}−90^{0 }= 70^{0}
Q10) In figure 16.139, if ∠ACB = 40^{0}, ∠DPB = 120^{0},find ∠CBD.
Solution:
We have,
∠ACB = 40^{0}; ∠DPB = 120^{0}
∴ ∠APB = ∠DCB = 40^{0} [Angleinsamesegment]
InΔPOB, by Angles umproperty ∠PDB+ ∠PBD+ ∠BPD = 180^{0}
⇒ 40^{0}+ ∠PBD+120^{0} = 180^{0}
⇒ ∠PBD = 180^{0}−40^{0}−120^{0}
⇒ ∠PBD = 20^{0}
∴ ∠CBD = 20^{0}
Q11) A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
We have,
Radius OA = Chord AB
= >OA = OB = AB
Then triangle OAB is an equilateral triangle.
∴ ∠AOB = 60^{0} [oneangle of equilateraltriangle]
by degree measure theorem
∠AOB = 2 ∠APB
⇒ 60^{0} = 2 ∠APB
⇒ ∠APB =
Now, ∠APB+ ∠AQB = 180^{0} [opposite Angles of cyclicquadrilateral]
⇒ 30^{0}+ ∠AQB = 180^{0}
⇒ ∠AQB = 180^{0}−30^{0} = 150^{0}.
Therefore, Angle by chord AB at minor arc = 150^{0}
Angle by chord AB at major arc = 30^{0}
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