Ex-16.4 (Part - 1), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1) In figure 16.120, O is the centre of the circle. If  ∠APB = 500,find ∠AOB and ∠OAB.

Solution:

∠APB = 50⁰

by degree measure theorem

∠AOB = 2 ∠APB
⇒ ∠APB = 2×50⁰ = 100⁰
since OA = OB [Radius of circle]
Then ∠OAB = ∠OBA [ Angles opposite to equalsides]
Let ∠OAB = x In ΔOAB, by Angles umproperty ∠OAB+ ∠OBA+ ∠AOB = 180⁰

= >x + x + 100⁰  = 180⁰

= >2x = 180⁰ – 100⁰

= >2x = 80⁰

= >x = 40⁰

∠OAB = ∠OBA = 40⁰

Q2) In figure 16.121, it is given that O is the centre of the circle and  ∠AOC = 150⁰.Find ∠ABC.

Solution:

∠AOC = 150⁰
∴ ∠AOC+reflex ∠AOC = 360⁰ [Complexangle]
⇒ 150⁰+reflex ∠AOC = 360⁰
⇒ reflex ∠AOC = 360⁰−150⁰
⇒ reflex ∠AOC = 210⁰
⇒ 2 ∠ABC = 210⁰ [ by degree measure theorem]
⇒ ∠ABC =  = 105⁰

Q3) In figure 16.22, O is the centre of the circle. Find  ∠BAC.

 Solution:

We have  ∠AOB = 80⁰

and  ∠AOC = 110⁰

Therefore,  ∠AOB+ ∠AOC+ ∠BOC = 360⁰ [Completeangle]

⇒ 80⁰+100⁰+ ∠BOC = 360⁰
⇒ ∠BOC = 360⁰−80⁰−110⁰
⇒ ∠BOC = 170⁰

by degree measure theorem

∠BOC = 2 ∠BAC
⇒ 170⁰ = 2 ∠BAC
⇒ ∠BAC =  = 85⁰

Q4) If O is the centre of the circle, find the value of x in each of the following figures.

(i)

Solution:

∠AOC = 135⁰
∴ ∠AOC+ ∠BOC = 180⁰ [Linearpair of Angles ]
⇒ 135⁰ + ∠BOC = 180⁰
⇒ ∠BOC = 180⁰−135⁰
⇒ ∠BOC = 45⁰

by degree measure theorem ∠BOC = 2 ∠CPB
⇒ 45⁰ = 2x

(ii)

Solution:

Wehave ∠ABC = 40⁰ ∠ACB = 90⁰ [Angle in semicircle]
In ΔABC, by Angles umproperty ∠CAB+ ∠ACB+ ∠ABC = 180⁰
⇒ ∠CAB+90⁰+40⁰ = 180⁰
⇒ ∠CAB = 180⁰−90⁰−40⁰
⇒ ∠CAB = 50⁰
Now, ∠CDB = ∠CAB [Angleissameinsegment]
⇒ x = 50⁰

(iii)

Solution:

Wehave ∠AOC = 120⁰ by degree measure theorem. ∠AOC = 2 ∠APC
⇒ 120⁰ = 2 ∠APC
⇒ ∠APC =  = 60⁰
∠APC+ ∠ABC = 180⁰ [Opposite Angles of cyclicquadrilaterals]
⇒ 60⁰+ ∠ABC = 180⁰
⇒ ∠ABC = 180⁰−60⁰
⇒ ∠ABC = 120⁰
∴ ∠ABC+ ∠DBC = 180⁰ [Linearpair of Angles ]
⇒ 12⁰+x = 180⁰
⇒ x = 180⁰−120⁰ = 60⁰

(iv)  

Solution:

Wehave ∠CBD = 65⁰
∴ ∠ABC+ ∠CBD = 180⁰ [Linearpair of Angles ]
⇒ ∠ABC = 65⁰ = 180⁰
⇒ ∠ABC = 180⁰−65⁰ = 115⁰
∴reflex ∠AOC = 2 ∠ABC [ by degree measure theorem]
⇒ x = 2×115⁰
⇒ x = 230⁰

(v)

Solution:

Wehave ∠OAB = 35⁰
Then, ∠OBA = ∠OAB = 35⁰ [ Angles opposite to equalradii]
InΔAOB, by Angles umproperty
⇒ ∠AOB+ ∠OAB+ ∠OBA = 180⁰
⇒ ∠AOB+35⁰+35⁰ = 180⁰
⇒ ∠AOB = 180⁰−35⁰−35⁰ = 110⁰
∴ ∠AOB+reflex ∠AOB = 360⁰ [Complexangle]
⇒ 110⁰+reflex ∠AOB = 360⁰
⇒ reflex ∠AOB = 360⁰−110⁰ = 250⁰ by degree measure theoremreflex ∠AOB = 2 ∠ACB
⇒ 250⁰ = 2x
⇒ x =  = 125⁰

(vi)

Solution:

Wehave ∠AOB = 60⁰ by degree measure theoremreflex ∠AOB = 2 ∠ACB
⇒ 60⁰ = 2 ∠ACB
⇒ ∠ACB = = 30⁰ [ Angles opposite to equalradii]
⇒ x = 30⁰.

(vii)

Solution:

Wehave ∠BAC = 50⁰ and ∠DBC = 70⁰
∴ ∠BDC = ∠BAC = 50⁰ [Angleinsamesegment]
InΔBDC, by Angles umproperty ∠BDC+ ∠BCD+ ∠DBC = 180⁰
⇒ 50⁰+x+70⁰ = 180⁰
⇒ x = 180⁰−50⁰−70⁰ = 60⁰

(viii)  

Solution:

Wehave, ∠DBO = 40⁰ and ∠DBC = 90⁰ [Angleinasemicircle]
⇒ ∠DBO+ ∠OBC = 90⁰
⇒ 40⁰+ ∠OBC = 90⁰
⇒ ∠OBC = 90⁰−40⁰ = 50⁰ by degree measure theorem
∠AOC = 2 ∠OBC
⇒ x = 2×50⁰ = 100⁰

(ix)

Solution:

InΔDAB, by Angles umproperty ∠ADB+ ∠DAB+ ∠ABD = 180⁰
⇒ 32⁰+ ∠DAB+50⁰ = 180⁰
⇒ ∠DAB = 180⁰−32⁰−50⁰
⇒ ∠DAB = 98⁰
Now, ∠OAB+ ∠DCB = 180⁰ [Opposite Angles of cyclicquadrilateral]
⇒ 98⁰ + x = 180⁰
⇒ x = 180⁰−98⁰ = 82⁰

(x)

Solution:

Wehave, ∠BAC = 35⁰
∠BDC = ∠BAC = 35⁰ [Angleinsamesegment]
InΔBCD, by Angles umproperty ∠BDC+ ∠BCD+ ∠DBC = 180⁰
⇒ 35⁰+x+65⁰ = 180⁰
⇒ x = 180⁰−35⁰−65⁰ = 80⁰

(xi)

Solution:

Wehave, ∠ABD = 40⁰
∠ACD = ∠ABD = 40⁰ [Angleinsamesegment]
InΔPCD, by Angles umproperty
∠PCD+ ∠CPO+ ∠PDC = 180⁰
⇒ 40⁰+110⁰+x = 180⁰
⇒ x = 180⁰−150⁰
⇒ x = 30⁰

 (xii)

Solution:

Giventhat, ∠BAC = 52⁰
Then ∠BDC = ∠BAC = 52⁰ [Angleinsamesegment]
SinceOD = OC
Then ∠ODC = ∠OCD [Oppositeangle to equalradii]
⇒ x = 52⁰