EXERCISE 24 B
Calculate the angles marked with letters. O is the center of the circle wherever given.
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°
Question 2
Sol :
∠C=2×40° (∠ by arc is twice at center, then on remaining circumference)
∠C=80°
Question 3
Sol :
100°=2x (∠ by arc is twice at center, then on remaining circumference)
$x=\frac{100}{2}$
x=50°
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°
Question 5
Sol :
∠a=35° (Angles in the same segment)
Question 6
Sol :
∠u=10° (Angles in the same segment)
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°
Question 8
Sol :
∠c+35°=90° (Angles in semi circle is right angle)
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°
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°
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°
Question 12
Sol :
∠e=110° (exterior angle=interior angle)
∠e+∠f=180° (linear pair)
110°+∠f=180°
∠f=180°-110°=70°
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°
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°
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°
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°
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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