S.chand books class 8 maths solution chapter 24 Circle exercise 24 B

EXERCISE 24 B

Calculate the angles marked with letters. O is the center of the circle wherever given.


Q1 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 1






Sol :

Angle made by an arc at the centre of a circle is twice the angle which this arc makes at the remaining part of the circumference.

120°=2×a

$a=\frac{120}{2}$

a=60°




Q2 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 2






Sol :

∠C=2×40° (∠ by arc is twice at center, then on remaining circumference)

∠C=80°




Q3 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 3






Sol :







100°=2x (∠ by arc is twice at center, then on remaining circumference)

$x=\frac{100}{2}$

x=50°




Q4 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 4


Sol :

∠a=2×40° (∠ by arc is twice at center, then on remaining circumference)

∠a=80°

In ΔOEF,

OE=OF (∵ they are radius)

So, ΔOEF is isosceles Δ

then

∠OEF=∠OFE=b


In ΔOEF,

∠EOF+∠OEF+∠OFE=180° (Angle sum property of Δ)

∠a+∠b+∠b=180° 

2∠b+∠a=180°

2∠b+80°=180°

2∠b=180°-80°

2∠b=100°

$\angle b=\frac{100}{2}$

∠b=50°




Q5 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 5









Sol :

∠a=35° (Angles in the same segment)

∠b=15°



Q6 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 6








Sol :

∠u=10° (Angles in the same segment)

∠w=30° (Angles in the same segment)



Q7 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 7






Sol :







∠a=40° (Angle in the same segment)

∠a+∠d+32°=180° (Angle sum property of Δ)

40°+∠d+32°=180°

∠d+72°=180°

∠d=180°-72°=108°



Q8 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 8







Sol :

∠c+35°=90° (Angles in semi circle is right angle)

∠c=90°-35°=55°



Q9 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 9






Sol :







∠x=90° (Angle in semi circle is right angle)

∠x+∠y+28°=180°

90°+∠y+28°=180°

∠y+118°=180°

∠y=180°-118°=62°

∠y=62°



Q10 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 10






Sol :







∠A=∠C (Angle in semi circle is 90°)

∠B=20° (Vertically Opposite Angle)

∠A+∠B+∠x=180°

90°+20°+∠x=180°

∠x=180°-110°=70°

Similarly, ∠y=70°



Q11 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 11






Sol :

∠a+95°=180° (Opposite angle of cyclic quadrilateral is supplementary)

∠a=180°-95°=85°

∠b+80°=180° (Opposite angle of cyclic quadrilateral is supplementary)

∠b=180°-80°=100°



Q12 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 12






Sol :

∠e=110° (exterior angle=interior angle)

∠e+∠f=180° (linear pair)

110°+∠f=180°

∠f=180°-110°=70°



Q13 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 13






Sol :









In cyclic quadrilateral PQRS,

∠a+104°=180° (Sum of opposite angle of quadrilateral is 180°)

∠a=180°-104°=76°


In ΔPQT,

∠P=90° (Angle in semi-circle is 90°)

∠P=90°

∠a+∠b=90°

76°+∠b=90°

∠b=90°-76°=14°




Q14 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 14






Sol :








∠p=42° (Angle in same segment are equal)

∠s+88°=180°

∠s=180°-88°=92°

∠s+∠q+∠p=180° (Angle sum property of triangle)

92°+∠q+42°=180°

∠q=180°-134°=46°



Q15 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 15







Sol :









In cyclic quadrilateral EHGF

∠a=50° (exterior ∠=interior opposite ∠)

∠b=2×∠a (Angle by an arc at centre is twice of remaining angle at circumference)

∠b=2×50°=100°


In ΔOEG (isosceles)

∠c+∠b+∠c=180° (Angle sum property of triangle)

2∠c+∠b=180°

2∠c=180°-100°

∠c$=\frac{80}{2}$

∠c=40°



Q16 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 16

(i) If ∠AOB=126° , what is ∠APB ?
(ii) If ∠OAB=29°, what is ∠APB ?
(iii) If ∠APB=62° , what is ∠OBA ?







Sol :









(i)

∠AOB=126° , then

∠AOB=2×∠APB (angle by arc at the centre twice of angle at remaining circumference)

126°=2×∠APB 

∠APB=63°


(ii)

In ΔOAB, OA=OB=(radii)

which makes them isosceles Δ

So, ∠OAB=∠OBA=29°

∠AOB+∠OAB+∠OBA=180° (Angle sum property of Δ)

∠AOB+29°+29°=180°

∠AOB=180°-58°=122°

∠AOB=2×∠APB=(Angle by arc at centre twice the remaining angle of circumference)

∠APB$=\frac{\angle AOB}{2}$

$=\frac{122}{2}$

∠APB=61°


(iii)

Given : ∠APB=62°

∠AOB=2×∠APB (Angle by arc at centre is twice as compare to remaining angle of circumference)

∠AOB=2×62°=124°


In ΔAOB is isosceles because OA=OB=radii

So, ∠OBA=∠OAB

∠AOB+∠OBA+∠OAB=180° (Angle sum property of Δ)

124°+∠OBA+∠OBA=180° (∠OBA=∠OAB)

2∠OBA=180°-124°

2∠OBA=56°

∠OBA$=\frac{56}{2}$

∠OBA=28°



Q17 | Ex-25B | Class 8 | S.Chand | Composite maths| Circle | Ch-25 | myhelper

Question 17

If BP||DC and ∠ACD=76° , find
(i) ∠ABD
(ii) ∠CAP
(iii) ∠BDC






Sol :








(i)

In cyclic quadrilateral , ABCD

∠ABD+76°=180°

∠ABD=180°-76°=104°


(ii)

BP||DC and AC is transversal

∠PAC=∠ACD (Alternate interior angle are equal)

∠PAC=76°


(iii)

∠PAC=∠BDC (if a  side is produced in a cyclic quadrilateral , the exterior angle so formed is equal to the interior opposite angle)

∠BDC=76°


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