Page 38 - mathsvol1ch1to3ans

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(i) 30 (ii) 135 ◦ (iii) −205 ◦ (iv) 150 ◦ (v) 330 .
Solution:
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◦
◦
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(i) 30 ≡ π (ii) 135 ≡ 3π (iii) −205 ≡ − 41π (iv) 150 ≡ 5π (v) 330 ≡ 11π
6 4 36 6 6
2. Find the degree measure corresponding to the following radian measures
π π 2π 7π 10π
(i) (ii) (iii) (iv) (v) .
3 9 5 3 9
Solution:
π π 2π 7π 10π
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(i) ≡ 60 (ii) ≡ 20 (iii) ≡ 72 (iv) ≡ 420 (v) ≡ 200 ◦
3 9 5 3 9
3. What must be the radius of a circular running path, around which an athlete must run 5 times in
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order to describe 1 km?
Circumference Circumference
Solution: Radius = ≈ .
2π 6.28
200
Since one round = 200 metres, radius = = 31.85 metres.
6.28
4. In a circle of diameter 40 cm, a chord is of length 20 cm. Find the length of the minor arc of the
chord.
Solution:
Diameter = 40cm. Length of the chord = 20cm. Radius = 20cm. Since radius =
length of chord = 20cm. Hence the formed triangle in the circle is equilateral
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triangle with each angle = 60 . We know that ` = rθ

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` = 20 × 60 × π
180 ◦
Figure 3.1
` = 20π .Thus length of the minor arc of the chord is 20π
3
3
5. Find the degree measure of the angle subtended at the centre of circle of radius 100 cm by an arc
of length 22cm.
Solution:We know that ` = rθ

` 22 0
◦
Hence θ = = = 0.22radians = 12 36 .
r 100
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6. What is the length of the arc intercepted by a central angle of measure 41 in a circle of radius
10 ft?
Solution: Since length of the arc = rθ = 41 × 0.017453 = 71.56feet
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7. If in two circles, arcs of the same length subtend angles 60 and 75 at the centre, find the ratio of
their radii.
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Solution:Let the length of the circle be `. Angle of the circle 1 = 60 . Angle of the circle 2 = 75 .
Let the radius be r 1 and r 2 .
◦
` = r 1 θ = r 1 × 60 × π = πr 1
180 3
◦
` = r 2 θ = r 2 × 75 × π = 5πr 2
180 12
Since length of the arcs are same, πr 1 : 5πr 2 , the ratio of the their radii is given by
3 12
r 1 : r 2 = 5 : 4
Figure 3.2