Let OX be the original line. The original line is also called initial line. Let a revolving line OP start from OX and revolve to reach into the position OP. The line OP is also called the terminal line. These lines form an angle XOP at the point O. The point O is called the vertex and we write \(\angle\)XOP to mean angle XOP.
If the direction of revolving line OX is clockwise, \(\angle\)XOQ is called the negative angle. But, if the direction of revolving line OP is anticlockwise, then \(\angle\)XOP is called the positive angle.
The following three systems are commonly used in the measurement of angles:
The system is also called British System. In this system, the unit of measurement is degree. So this system is also known as degree system. A right angle is divided into 90 equal parts and each part is called a degree. A degree is divided into 60 equal parts and each part is called a minute. A minute is also divided into 60 equal parts and each part is called a second. Therefore, we have
60 seconds = 1 minutes (60" =1')
60 minutes = 1 degree (60' = 1°)
90 degrees = 1 right angle
The degree, minute the second are denoted by (°), (') and (")
This system of measurement is also called the French system. In this system, the unit of measurement is grade. So, this system is also known as grade system. A right angle is divided into 100 equal parts, each part is called a grade. A grade is divided into 100 equal part, each part is called a minute. A minute is also divided into 100 equal parts, each part is called a second. Therefore, we have,
100 seconds = 1 minute (100" = 1')
100 minutes = 1 grade (100' = 1g)
100 grades = 1 right angle
The grade, minute and second are respectively denoted by (g), (') and (").
Draw a circle with centre O and radius OP. Take any point B on the circumference such that arc PQ is equal in length to the radius OP. Join QO. The angle POQ, so formed , is said to be 1 radian. A radian is denoted by (c). So, 1 radian is written as 1c. Here, ∠POQ = 1c and arc PQ = r.
An angle subtended by an arc equal in length to the radius, at the centre of a circle is defined as a radian.
Let OP = OQ = r. Produce PO to the point R on the circumference of the circle. Then PR is called the diameter of the circle.
Here, circumference of the circle = 2πr
∴ arc PQR =πr
We know angles subtended by different arcs at the centre of a circle are proportional to the corresponding arcs .
So, \(\frac{∠POQ}{∠POR}\) = \(\frac{arc PQ}{arc PR}\)
or, \(\frac{1°}{180°}\) = \(\frac{r}{πr}\)
∴ 180° =π^{c}
Polygon is a closed plane figure having three or more than three line segments. Triangles, quadrilateral, nonagon, octagon etc. are the examples of a polygon. A polygon having all sides equal in length is called a regular polygon. A regular polygon has the same measures of interior angles.
In regular polygons of sides n, each interior angle is θ = \(\frac{(n-2) × 180°}{n}\)
Similarly,
The exterior angle is a side of a regular polygon where an angle between any side of a shape and a line extended from the next side.
Exterior angle of polygon (Φ) = \(\frac{360°}{n}\)
The radian is a constant angle
Proof: Let O be the center of the circle and AB be the diameter. Let C be a point on the circumference such that OB = arc BC = r where r is the circle. Then, by definition,∠BOC = 1c.
We know, circumference of a circle = 2πr
∴ Length of arc BCA = \(\frac{1}{2}\)× 2πr =πr.
Now, angles subtended by different arcs at the center o0f a circle are proportional to the corresponding arcs.
Therefore we have
\(\frac{∠BOC}{∠AOB}\) = \(\frac{arc \: BC}{arc \: BCA}\)
or, \(\frac{1^c}{180^o}\) = \(\frac{r}{πr}\)
or, 1c = \(\frac{180^o}{π}\) which is constant
Hence, the radian is a constant angle.
The circular measure of an angle subtended by an arc of a circle at the center is expressed by the ratio of the arc to the radius of a circle.
i.e. angle subtended at the centre of a circle = \(\frac{length \: of \: the \: arc \: of \: the \: circle}{radius \: of \: the \: circle}\)
Proof: Let O be the center of a circle and r be its radius. Take any two points A and B on the circle such that arc AB = r.
Then, by the definition,∠AOB = 1c.
Let C be any other point on the circle such that arc ABC = \ and the angle subtended by arc ABC at the center of the circle be θc.
We know angles subtended by different arcs at the center of a circle are proportional to the corresponding arcs.
Therefore, we have
\(\frac{Arc \: ABC}{Arc \: AB}\) = \(\frac{∠AOB}{∠AOC}\)
or, \(\frac{l}{r}\) = \(\frac{θ^c}{1^c}\)
∴θ^{c} = \(\frac{l}{r}\)
Thus, angle at the center = \(\frac{length \: of \: the \: arc \: of \: the \: circle}{radius \: of \: the \: circle}\)
From this theorem, we have following three relations:
When the central angleθ^{c} becomes 2π, then the corresponding arc l becomes circumference of the circle. Then, by formula (ii) we have
l = rθ
or, circumference of circle = r× 2π = 2πr.
In a clock, the hands of the clock always rotate is the clock-wise direction. The minute hand, the hour hand and the second-hand together show us the time. All the hands of the clock are moving always but only the change in second hand can be observed as it makes a remarkable change within seconds. But all the hands of the clock are always in the state of change.
The second hand,
The second-hand makes 1 complete rotation in 1 minute.
i.e. In 1 minute = 60 sec, the second-hand turns 360.
\(\therefore\) In 1 second, the second hand turns = \(\frac{360°}{60}\) = 6°
The minute hand,
The minute hand makes 1 complete rotation in 1 hour.
i.e. In 1 minute = 60 min, the minute hand turns 360°.
\(\therefore\) In 1 minute, the minute hand turns \(\frac{360°}{60}\) = 6°
The hour hand,
The hour hand makes 1 complete rotation in 12 hours.
i.e. In 12 hours, the minute hand turns 360°.
\(\therefore\) In 1 hour, the minute hand turns = \(\frac{360°}{12}\) = 30°.
→ Sexagesimal system (British System)
→ Cetesimal system (French System)
→ Radian system (circular system)
Soln:
Here, 27^{o} = 27× 60' = 1620'
27^{o}38' = (1620 + 38)' = 1658'
Then, 1658' = (1658 × 60)" = 99480"
∴ 27^{o}38'42" = (99480 + 42)" = 99522". Ans.
Soln:
Here, 60^{o} = 60× 60' = 600'
3600' = (3600× 60") = 216000"
∴ 60^{o} = 216000" Ans.
Soln:
Here, 10" =(\(\frac{12}{60}\))
10' 12" =(10 +\(\frac{12}{60}\))' = (10 + 0.2)' = 10.2'
10.2' =(\(\frac{10.2}{60}\))^{o} = (\()\frac{102}{600}\)^{o} = 0.17^{o}
∴ 20^{o} 10' 12" = (20 + 0.17)^{o} = 20.17^{o} . Ans.
Soln:
Here, 10" = (\(\frac{10}{60}\))' = (\(\frac{1}{6}\))'
10' 10" = (10 + \(\frac{1}{60}\)) =(\(\frac{60 + 1}{6}\))' =\(\frac{61'}{6}\)
\(\frac{61'}{6}\) =(\(\frac{61}{6}\)× \(\frac{1}{60}\))^{o} =(\(\frac{61}{360}\))^{o} = 0.1694^{o}
∴ 20^{o} 10' 10" = (20 + 0.1694)^{o} = 20.1694^{o}. Ans.
Soln:
Here, 43" =(\(\frac{43}{6}\))'
13'43" =(13 + \(\frac{43}{60}\))' =(\(\frac{823}{60}\))'
(\(\frac{823}{60}\))' =(\(\frac{823}{60}\)×\(\frac{1}{60}\))^{o} = 0.2286^{o}
∴ 57^{o} 13' 43" = ( 57 + 0.2286)^{o} = 57.2286^{o} Ans.
Soln:
Here, 30" =(\(\frac{30}{60}\))' =(\(\frac{1}{2}\))'
15'30" = (15 + \(\frac{1}{2}\))' =(\(\frac{31}{2}\))'
(\(\frac{31}{2}\))' =(\(\frac{31}{2}\)×\(\frac{1}{60}\))^{o} = 0.2583^{o}
∴ 60^{o} 15' 30" = ( 60 + 0.2583)^{o} = 60.2583^{o} Ans.
Soln:
Here, 25" = (\(\frac{25}{100}\))' =(\(\frac{1}{4}\))'
14' 25" = ( 14 + \(\frac{1}{4}\))' =(\(\frac{57}{4}\))'
(\(\frac{57}{4}\))' =(\(\frac{57}{4}\) × \(\frac{1}{100}\))^{g} =(\(\frac{57}{400}\))^{g} = 0.1425^{g}
∴ 6^{g} 14' 25" = 6 + 0.1425^{g} = 6.1425^{g} Ans.
Soln:
Here, 85" = (\(\frac{85}{100}\))' = (\(\frac{17}{20}\))'
2'85" = ( 2 + \(\frac{17}{20}\))' = (\(\frac{57}{20}\))'
(\(\frac{57}{20}\))' =(\(\frac{57}{20}\)×\(\frac{1}{100}\))^{g} =(\(\frac{57}{2000}\))^{g} = 0.0285^{g}
∴ 80^{g} 2' 85" = ( 80 + 0.0285)^{g} = 80.0285^{g}. Ans.
Soln:
Here, 90" = (\(\frac{90}{100}\))' = (\(\frac{9}{10}\))'
5' 90" = ( 5 + \(\frac{9}{10}\))' = (\(\frac{59}{10}\))'
(\(\frac{59}{10}\))' =(\(\frac{59}{10}\)×\(\frac{1}{100}\))^{g} = 0.059^{g}
∴75^{g} 5' 90" = (75 + 0.059)^{g} = 75.059^{g}. Ans.
Soln:
Here, 70" = (\(\frac{70}{100}\))' =(\(\frac{7}{10}\))'
(\(\frac{7}{10}\))' =(\(\frac{7}{10}\)×\(\frac{1}{100}\))^{g} = (\(\frac{7}{1000}\))^{g} = 0.007^{g}
∴ 65^{g} 70" = ( 65 + 0.007)^{g} = 65.007^{g} Ans.
Soln:
Here, 25^{g} 26' 23" = 25.2623^{g} [ 25 +\(\frac{26}{100}\) + \(\frac{23}{10000}\) = 25+0.26+0.0023]
Changing into grade
25.2623^{g} =(25.2623× \(\frac{9}{10}\))^{o}
= 22.73607^{o} = 22^{o} 44' 9.85". Ans.
Soln:
Here,30^{g} 29' 19.75" = 30.291975^{g} [ 30 + \(\frac{29}{100}\) + \(\frac{19.75}{10000}\) = 30 + 0.29 + 0.001975]
30.29195^{g} = (30.291975× \(\frac{9}{10}\))^{o}
= 27.262777^{o} = 27^{o} 15' 46" Ans.
Soln:
Here,26^{o}46' = (26 +\(\frac{46}{60}\))^{o} = ( 26 + 0.766)^{o} = 26.7666^{o}
26.7666^{o} = ( 26.7666×\(\frac{10}{9}\))^{g} = 29.74066^{g}
= 29^{g} 74' 6.6" Ans.
Solution:
Let the no. of sides of the regular polygon be n.
Then, exterior angle = \(\frac{}{}\) × interior angle.
i.e. \(\frac{360}{n}\) = \(\frac{1}{4}\) × \(\frac{(n - 2) × 180°)}{n}\)
or, \(\frac{360}{n}\) = \(\frac{45(n - 2)}{n}\)
or, n - 2 = \(\frac{360}{45}\)
or, n - 2 = 8
\(\therefore\) n = 10
Solution
When the clock strikes half past three.
The hour hand at 3 and minute hand at 6.
So, for real calculation,
The hour hand moves 30 minutes away from 3.
So, In 1 hour, the hour hand makes 30°.
\(\therefore\) In \(\frac{}{}\) hour, the hour hand makes 30 × \(\frac{1°}{2}\) = 15°
Also, the angle between the hour and minute hand when the hand is at 3 and 6 is 90°.
So, the angle between the two hands at half past three 90° - 15° = 75°.
The radius of a circle is 5.6 cm.Find the angle in degrees subtended by an area of 22 cm arc at the centre of the circle.
The radius of a circle is 7 cm.Find the angle in degrees subtended by a arc of 3(frac{2}{3}) cm at the centre of the circle.
30°
40°
20°
10°
Find the angle in degrees subtended by an arc of length 44cm at the centre of a circle of radius 21 cm
120°
119°
220°
126°
The radius of a circle is 21 cm.An arc of the circle subtends an angle of 60° at the centre.Find the length of the arc.
12 cm
19 cm
22 cm
10 cm
If the sum of degrees, minutes and seconds of any angle is 32949sec, find the angle in degree.
If the sum of degrees, minutes and seconds of any angle is 164745sec, find the angle in degree.
A cow is tied to a pole by a rope of length 12 m. If a cow grazes such that it describes a circle of radius 12 m. how far will it have moved when the rope traces an angle of 60° at the pole?
12.57 m
2 m
8 m
5 m
A horse is tied to a pole by a rope of length 14 m. If a horse grazes such that it describes a circle of radius 14 m. how far will it have moved when the rope traces an angle of 45° at the pole?
11 m
13 m
12 m
14 m
If the sum of the number of the degrees and the numbers of grades of an angle is 152, find the angle in the degree.
42°
32°
22°
72°
If the sum of the number of the degrees and grades of ∠A is 95. find the angle in the degree.
45°
75°
65°
55°
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