Ganit Prakash - Class-X - Trigonometry
Let us work out 20 Measurement of Angle
📘 Exercise 20 Solutions (Q1 - Q13)
Trigonometry(i) 832′ (ii) 6312″ (iii) 375″ (iv) \( 27\frac{1}{12}^\circ \) (v) \( 72.04^\circ \)
Explanation:
We know that \( 1^\circ = 60' \) (minutes) and \( 1' = 60'' \) (seconds).
(i) 832′
Divide by 60 to convert minutes to degrees:
\[\begin{array}{l} 832' = \left(\frac{832}{60}\right)^\circ \\ = 13^\circ \text{ with a remainder of } 52' \\ = \mathbf{13^\circ \, 52'} \end{array}\](ii) 6312″
Divide by 60 to convert seconds to minutes:
\[\begin{array}{l} 6312'' = \left(\frac{6312}{60}\right)' \\ = 105' \text{ with a remainder of } 12'' \end{array}\]Now, divide 105' by 60 to get degrees:
\[\begin{array}{l} 105' = \left(\frac{105}{60}\right)^\circ \\ = 1^\circ \text{ with a remainder of } 45' \\ \text{So, } 6312'' = \mathbf{1^\circ \, 45' \, 12''} \end{array}\](iii) 375″
Divide by 60 to convert seconds to minutes:
\[\begin{array}{l} 375'' = \left(\frac{375}{60}\right)' \\ = 6' \text{ with a remainder of } 15'' \\ = \mathbf{6' \, 15''} \text{ (or } 0^\circ \, 6' \, 15''\text{)} \end{array}\](iv) \( 27\frac{1}{12}^\circ \)
Separate the whole degree and the fractional degree:
\[\begin{array}{l} 27\frac{1}{12}^\circ = 27^\circ + \left(\frac{1}{12}\right)^\circ \\ = 27^\circ + \left(\frac{1}{12} \times 60\right)' \\ = 27^\circ + 5' \\ = \mathbf{27^\circ \, 5'} \end{array}\](v) \( 72.04^\circ \)
Separate the whole degree and the decimal part:
\[\begin{array}{l} 72.04^\circ = 72^\circ + 0.04^\circ \\ = 72^\circ + (0.04 \times 60)' \\ = 72^\circ + 2.4' \\ = 72^\circ + 2' + 0.4' \\ = 72^\circ + 2' + (0.4 \times 60)'' \\ = 72^\circ + 2' + 24'' \\ = \mathbf{72^\circ \, 2' \, 24''} \end{array}\](i) 60° (ii) 135° (iii) –150° (iv) 72° (v) 22°30′ (vi) –62°30′ (vii) 52° 52′ 30″ (viii) 40° 16′ 24″
Explanation:
To convert degrees to circular measure (radians), we multiply by \( \frac{\pi}{180^\circ} \).
(i) 60° = \( 60 \times \frac{\pi}{180} = \mathbf{\frac{\pi}{3}} \text{ or } \mathbf{\frac{\pi^c}{3}} \)
(ii) 135° = \( 135 \times \frac{\pi}{180} = \frac{3\pi}{4} = \mathbf{\frac{3\pi^c}{4}} \)
(iii) –150° = \( -150 \times \frac{\pi}{180} = -\frac{5\pi}{6} = \mathbf{-\frac{5\pi^c}{6}} \)
(iv) 72° = \( 72 \times \frac{\pi}{180} = \frac{2\pi}{5} = \mathbf{\frac{2\pi^c}{5}} \)
(v) 22°30′
\[\begin{array}{l} = 22^\circ + \left(\frac{30}{60}\right)^\circ = 22^\circ + \frac{1^\circ}{2} = \left(\frac{45}{2}\right)^\circ \\ \text{Circular value } = \frac{45}{2} \times \frac{\pi}{180} = \mathbf{\frac{\pi^c}{8}} \end{array}\](vi) –62°30′
\[\begin{array}{l} = -\left(62^\circ + \left(\frac{30}{60}\right)^\circ\right) = -\left(62 + \frac{1}{2}\right)^\circ = -\left(\frac{125}{2}\right)^\circ \\ \text{Circular value } = -\frac{125}{2} \times \frac{\pi}{180} = -\frac{25\pi}{2 \times 36} = \mathbf{-\frac{25\pi^c}{72}} \end{array}\](vii) 52° 52′ 30″
Convert everything to degrees first:
\[\begin{array}{l} 52' \, 30'' = 52' + \left(\frac{30}{60}\right)' = 52.5' = \left(\frac{105}{2}\right)' \\ \text{Now, } 52^\circ \, \left(\frac{105}{2}\right)' = 52^\circ + \left(\frac{105}{2 \times 60}\right)^\circ \\ = 52^\circ + \left(\frac{7}{8}\right)^\circ = \left(\frac{416 + 7}{8}\right)^\circ = \left(\frac{423}{8}\right)^\circ \\ \text{Circular value } = \frac{423}{8} \times \frac{\pi}{180} = \frac{47\pi}{8 \times 20} = \mathbf{\frac{47\pi^c}{160}} \end{array}\](viii) 40° 16′ 24″
\[\begin{array}{l} 16' \, 24'' = 16' + \left(\frac{24}{60}\right)' = 16' + \left(\frac{2}{5}\right)' = \left(\frac{82}{5}\right)' \\ \text{Now, } 40^\circ \, \left(\frac{82}{5}\right)' = 40^\circ + \left(\frac{82}{5 \times 60}\right)^\circ \\ = 40^\circ + \left(\frac{41}{150}\right)^\circ = \left(\frac{6000 + 41}{150}\right)^\circ = \left(\frac{6041}{150}\right)^\circ \\ \text{Circular value } = \frac{6041}{150} \times \frac{\pi}{180} = \mathbf{\frac{6041\pi^c}{27000}} \end{array}\]Explanation:
BCD is a straight line, so \( \angle ACB \) and \( \angle ACD \) are supplementary.
\[\begin{array}{l} \angle ACB + \angle ACD = 180^\circ \\ \Rightarrow \angle ACB + 144^\circ = 180^\circ \\ \Rightarrow \angle ACB = 180^\circ - 144^\circ = 36^\circ \end{array}\]In \( \Delta ABC \), it is given that AC = BC. This means it is an isosceles triangle, and the angles opposite to the equal sides are equal.
Therefore, \( \angle CAB = \angle ABC \).
We know the sum of angles in a triangle is \( 180^\circ \):
\[\begin{array}{l} \angle CAB + \angle ABC + \angle ACB = 180^\circ \\ \Rightarrow 2\angle CAB + 36^\circ = 180^\circ \\ \Rightarrow 2\angle CAB = 180^\circ - 36^\circ = 144^\circ \\ \Rightarrow \angle CAB = 72^\circ \end{array}\]So, the angles are \( \angle ACB = 36^\circ \), \( \angle CAB = 72^\circ \), and \( \angle ABC = 72^\circ \).
Now, let's find their circular values:
\[\begin{array}{l} \text{Circular value of } 36^\circ = 36 \times \frac{\pi}{180} = \mathbf{\frac{\pi^c}{5}} \\ \text{Circular value of } 72^\circ = 72 \times \frac{\pi}{180} = \mathbf{\frac{2\pi^c}{5}} \end{array}\]Answer: The circular values are \( \frac{\pi^c}{5}, \frac{2\pi^c}{5}, \frac{2\pi^c}{5} \).
Explanation:
Let the two acute angles of the right-angled triangle be \( x \) and \( y \), where \( x > y \).
Since it's a right-angled triangle, the sum of the two acute angles is \( 90^\circ \).
\[\begin{array}{l} x + y = 90^\circ \quad \text{--- (1)} \end{array}\]We are given the difference in circular measure is \( \frac{2\pi}{5} \). Let's convert this to sexagesimal (degrees):
\[\begin{array}{l} \frac{2\pi}{5} = \frac{2 \times 180^\circ}{5} = 2 \times 36^\circ = 72^\circ \end{array}\]So, the difference is \( 72^\circ \):
\[\begin{array}{l} x - y = 72^\circ \quad \text{--- (2)} \end{array}\]Now, add equation (1) and equation (2):
\[\begin{array}{l} (x + y) + (x - y) = 90^\circ + 72^\circ \\ \Rightarrow 2x = 162^\circ \\ \Rightarrow x = 81^\circ \end{array}\]Substitute the value of \( x \) in equation (1):
\[\begin{array}{l} 81^\circ + y = 90^\circ \\ \Rightarrow y = 90^\circ - 81^\circ = 9^\circ \end{array}\]Answer: The sexagesimal values of the two angles are \( 81^\circ \) and \( 9^\circ \).
Explanation:
Let the three angles of the triangle be \( A, B, \) and \( C \).
Given: \( A = 65^\circ \)
Given: \( B = \frac{\pi}{12} \). Let's convert this to degrees:
\[\begin{array}{l} B = \frac{180^\circ}{12} = 15^\circ \end{array}\]We know the sum of angles in a triangle is \( 180^\circ \):
\[\begin{array}{l} A + B + C = 180^\circ \\ \Rightarrow 65^\circ + 15^\circ + C = 180^\circ \\ \Rightarrow 80^\circ + C = 180^\circ \\ \Rightarrow C = 180^\circ - 80^\circ = 100^\circ \end{array}\]So, the sexagesimal value of the third angle is \( 100^\circ \).
Now, let's find its circular value:
\[\begin{array}{l} \text{Circular value} = 100 \times \frac{\pi}{180} = \frac{10\pi}{18} = \frac{5\pi}{9} \text{ or } \frac{5\pi^c}{9} \end{array}\]Answer: The sexagesimal value is \( 100^\circ \) and the circular value is \( \frac{5\pi^c}{9} \).
Explanation:
Let the two angles be \( x \) and \( y \), with \( x > y \).
Given sum: \( x + y = 135^\circ \) --- (1)
Given difference: \( x - y = \frac{\pi}{12} \). Convert this to degrees:
\[\begin{array}{l} x - y = \frac{180^\circ}{12} = 15^\circ \quad \text{--- (2)} \end{array}\]Add equation (1) and (2):
\[\begin{array}{l} (x + y) + (x - y) = 135^\circ + 15^\circ \\ \Rightarrow 2x = 150^\circ \\ \Rightarrow x = 75^\circ \end{array}\]Substitute the value of \( x \) in equation (1):
\[\begin{array}{l} 75^\circ + y = 135^\circ \\ \Rightarrow y = 135^\circ - 75^\circ = 60^\circ \end{array}\]So, the sexagesimal values are \( 75^\circ \) and \( 60^\circ \).
Now, let's find their circular values:
\[\begin{array}{l} \text{Circular value of } 75^\circ = 75 \times \frac{\pi}{180} = \frac{5\pi}{12} = \mathbf{\frac{5\pi^c}{12}} \\ \text{Circular value of } 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3} = \mathbf{\frac{\pi^c}{3}} \end{array}\]Explanation:
Let the three angles of the triangle be \( 2x \), \( 3x \), and \( 4x \).
We know the sum of angles in a triangle is \( 180^\circ \):
\[\begin{array}{l} 2x + 3x + 4x = 180^\circ \\ \Rightarrow 9x = 180^\circ \\ \Rightarrow x = 20^\circ \end{array}\]The greatest angle is \( 4x \):
\[\begin{array}{l} \text{Greatest angle } = 4 \times 20^\circ = 80^\circ \end{array}\]Now, we convert the greatest angle to its circular value:
\[\begin{array}{l} \text{Circular value } = 80 \times \frac{\pi}{180} \\ = \frac{8\pi}{18} \\ = \frac{4\pi}{9} \text{ or } \frac{4\pi^c}{9} \end{array}\]Answer: The circular value of the greatest angle is \( \frac{4\pi^c}{9} \).
Explanation:
We know the formula relating the arc length (\( s \)), radius (\( r \)), and the central angle in radians (\( \theta \)):
\[\begin{array}{l} \theta = \frac{s}{r} \end{array}\]Given:
Arc length \( (s) = 5.5 \text{ cm} \)
Radius \( (r) = 28 \text{ cm} \)
Substituting the values:
\[\begin{array}{l} \theta = \frac{5.5}{28} \\ \theta = \frac{55}{280} \quad \text{[Multiplying numerator and denominator by 10]} \\ \theta = \frac{11}{56} \text{ radians} \end{array}\]Answer: The circular value of the angle subtended is \( \frac{11}{56} \) radians (or \( \frac{11^c}{56} \)).
Explanation:
Let the two angles be \( 5x \) and \( 2x \) according to the given ratio.
We are given that the second angle is \( 30^\circ \).
\[\begin{array}{l} 2x = 30^\circ \\ \Rightarrow x = \frac{30^\circ}{2} = 15^\circ \end{array}\]Now, we can find the sexagesimal value of the first angle:
\[\begin{array}{l} \text{First angle} = 5x \\ = 5 \times 15^\circ = 75^\circ \end{array}\]To find the circular value of the first angle, we multiply by \( \frac{\pi}{180^\circ} \):
\[\begin{array}{l} \text{Circular value} = 75 \times \frac{\pi}{180} \\ = \frac{75\pi}{180} \\ = \frac{5\pi}{12} \text{ radians} \quad \text{[Dividing by 15]} \end{array}\]Answer: The sexagesimal value of the first angle is \( 75^\circ \) and its circular value is \( \frac{5\pi^c}{12} \).
Explanation:
The given angle is \( -5\frac{1}{12}\pi \).
1. Direction: Since the angle is negative, the ray has rotated in the clockwise direction.
2. Rotations: Let's break down the angle. We know that one complete rotation is \( 2\pi \) radians (or \( 360^\circ \)).
\[\begin{array}{l} 5\frac{1}{12}\pi = \frac{61\pi}{12} \end{array}\]We divide \( 61\pi \) by \( 12 \) to find the number of full \( 2\pi \) rotations:
\[\begin{array}{l} \frac{61\pi}{12} = \frac{48\pi + 13\pi}{12} \\ = \frac{48\pi}{12} + \frac{13\pi}{12} \\ = 4\pi + \frac{13\pi}{12} \\ = 2 \times (2\pi) + \frac{13\pi}{12} \end{array}\]Here, \( 4\pi \) represents 2 complete rotations (since \( 2 \times 2\pi = 4\pi \)).
3. Remaining Angle: After 2 complete rotations, the extra angle produced is \( \frac{13\pi}{12} \).
We can also express this extra angle in degrees:
\[\begin{array}{l} \frac{13\pi}{12} \text{ radians} = \frac{13 \times 180^\circ}{12} = 13 \times 15^\circ = 195^\circ \end{array}\]Answer: The ray has rotated in the clockwise direction. It has completed 2 full rotations and thereafter produced an angle of \( \frac{13\pi^c}{12} \) (or \( 195^\circ \)).
Explanation:
Given \( \Delta ABC \) is isosceles, and the included angle between the equal sides is \( \angle ABC = 45^\circ \). This means sides \( AB = BC \).
Therefore, the base angles are equal: \( \angle BAC = \angle BCA \).
Sum of angles in a triangle is \( 180^\circ \):
\[\begin{array}{l} \angle BAC + \angle BCA + \angle ABC = 180^\circ \\ 2\angle BAC + 45^\circ = 180^\circ \\ 2\angle BAC = 135^\circ \\ \angle BAC = \angle BCA = 67.5^\circ \end{array}\]BD is the angle bisector of \( \angle ABC \). So, it divides \( \angle ABC \) into two equal halves:
\[\begin{array}{l} \angle ABD = \angle CBD = \frac{45^\circ}{2} = 22.5^\circ \end{array}\]Now, let's find the circular value for each required angle:
1. \( \angle ABD \) and \( \angle CBD \):
\[\begin{array}{l} 22.5^\circ = \frac{45^\circ}{2} \\ \text{Circular value} = \frac{45}{2} \times \frac{\pi}{180} = \mathbf{\frac{\pi^c}{8}} \end{array}\]2. \( \angle BAD \) (which is \( \angle BAC \)) and \( \angle BCD \) (which is \( \angle BCA \)):
\[\begin{array}{l} 67.5^\circ = \frac{135^\circ}{2} \\ \text{Circular value} = \frac{135}{2} \times \frac{\pi}{180} = \frac{3\pi}{2 \times 4} = \mathbf{\frac{3\pi^c}{8}} \end{array}\]Answer: The circular values are: \( \angle ABD = \mathbf{\frac{\pi^c}{8}} \), \( \angle BAD = \mathbf{\frac{3\pi^c}{8}} \), \( \angle CBD = \mathbf{\frac{\pi^c}{8}} \), and \( \angle BCD = \mathbf{\frac{3\pi^c}{8}} \).
Explanation:
Given \( \Delta ABC \) is an equilateral triangle, so all its sides are equal (\( AB = BC = AC \)) and all its angles are \( 60^\circ \).
The base BC is extended to E such that \( CE = BC \).
Since BCE is a straight line, \( \angle ACE \) and \( \angle ACB \) form a linear pair.
\[\begin{array}{l} \angle ACE = 180^\circ - \angle ACB \\ \angle ACE = 180^\circ - 60^\circ = 120^\circ \end{array}\]In \( \Delta ACE \), we know \( AC = BC \) (sides of equilateral triangle) and it is given \( CE = BC \). Therefore, \( AC = CE \).
This makes \( \Delta ACE \) an isosceles triangle. So, the angles opposite to the equal sides are equal: \( \angle CAE = \angle CEA \).
Sum of angles in \( \Delta ACE \) is \( 180^\circ \):
\[\begin{array}{l} \angle CAE + \angle CEA + \angle ACE = 180^\circ \\ 2\angle CAE + 120^\circ = 180^\circ \\ 2\angle CAE = 60^\circ \\ \angle CAE = \angle CEA = 30^\circ \end{array}\]Now, we convert the angles of \( \Delta ACE \) (\( 120^\circ, 30^\circ, 30^\circ \)) to circular values:
\[\begin{array}{l} \text{Circular value of } \angle ACE (120^\circ) = 120 \times \frac{\pi}{180} = \mathbf{\frac{2\pi^c}{3}} \\ \text{Circular value of } \angle CAE (30^\circ) = 30 \times \frac{\pi}{180} = \mathbf{\frac{\pi^c}{6}} \\ \text{Circular value of } \angle AEC (30^\circ) = 30 \times \frac{\pi}{180} = \mathbf{\frac{\pi^c}{6}} \end{array}\]Answer: The circular values of the angles of \( \Delta ACE \) are \( \frac{2\pi^c}{3} \), \( \frac{\pi^c}{6} \), and \( \frac{\pi^c}{6} \).
Explanation:
Let the four angles of the quadrilateral be A, B, C, and D.
First, let's convert all the given angles to degrees (sexagesimal system):
\[\begin{array}{l} A = \frac{\pi}{3} = \frac{180^\circ}{3} = 60^\circ \\ B = \frac{5\pi}{6} = \frac{5 \times 180^\circ}{6} = 5 \times 30^\circ = 150^\circ \\ C = 90^\circ \end{array}\]We know that the sum of the four angles of a quadrilateral is always \( 360^\circ \).
\[\begin{array}{l} A + B + C + D = 360^\circ \\ 60^\circ + 150^\circ + 90^\circ + D = 360^\circ \\ 300^\circ + D = 360^\circ \\ D = 360^\circ - 300^\circ \\ D = 60^\circ \end{array}\]So, the sexagesimal value of the fourth angle is \( 60^\circ \).
Now, let's find its circular value:
\[\begin{array}{l} \text{Circular value} = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians} \end{array}\]Answer: The sexagesimal value of the fourth angle is \( 60^\circ \) and its circular value is \( \frac{\pi^c}{3} \).
📘 14. Very short Answer : (V.S.A.)
(A) M.C.Q.Explanation:
In exactly 1 hour (60 minutes), the minute hand of a clock completes one full revolution around the center.
One complete revolution subtends an angle of \( 360^\circ \) at the center.
Converting \( 360^\circ \) to circular measure (radians):
\[\begin{array}{l} 360^\circ \times \frac{\pi}{180^\circ} = 2\pi \text{ radian} \end{array}\]Correct Option: (d) \( 2\pi \) radian
Explanation:
To convert from circular measure (radians) to sexagesimal measure (degrees), we substitute \( \pi \) with \( 180^\circ \) or multiply the radian value by \( \frac{180^\circ}{\pi} \).
\[\begin{array}{l} \frac{\pi}{6} \text{ radian} = \frac{180^\circ}{6} = 30^\circ \end{array}\]Correct Option: (d) 30°
Explanation:
A regular hexagon is a polygon with \( n = 6 \) equal sides.
The formula for finding the value of each internal angle of a regular polygon with \( n \) sides is \( \frac{(n - 2) \times 180^\circ}{n} \).
\[\begin{array}{l} \text{Internal angle} = \frac{(6 - 2) \times 180^\circ}{6} \\ = \frac{4 \times 180^\circ}{6} \\ = 4 \times 30^\circ = 120^\circ \end{array}\]Converting \( 120^\circ \) to circular measure (radians):
\[\begin{array}{l} 120 \times \frac{\pi}{180} = \frac{12\pi}{18} = \frac{2\pi}{3} \end{array}\]Correct Option: (b) \( \frac{2\pi}{3} \)
Explanation:
The fundamental mathematical relationship \( S = r\theta \) (where \( S \) is the arc length and \( r \) is the radius) holds true only when the central angle \( \theta \) is measured in radians.
The radian is the unit of angle measure in the circular system.
Correct Option: (b) circular system
Explanation:
A key property of a cyclic quadrilateral is that the sum of any pair of opposite angles is \( 180^\circ \) (they are supplementary).
Since \( \angle A \) and \( \angle C \) form a pair of opposite angles:
\[\begin{array}{l} \angle A + \angle C = 180^\circ \\ 120^\circ + \angle C = 180^\circ \\ \Rightarrow \angle C = 180^\circ - 120^\circ = 60^\circ \end{array}\]Converting \( 60^\circ \) to circular measure:
\[\begin{array}{l} 60 \times \frac{\pi}{180} = \frac{\pi}{3} \end{array}\]Correct Option: (a) \( \frac{\pi}{3} \)
📘 14. Very short Answer : (V.S.A.)
(B) True or FalseThe angle formed by rotating a ray about its end point in anticlockwise direction is positive.
Explanation:
According to the standard mathematical sign convention in trigonometry, an angle generated by rotating a ray in the anti-clockwise (counter-clockwise) direction from its initial position is considered a positive angle.
Answer: True
The angle formed by rotating a ray about its end point in clockwise direction is positive.
Explanation:
As per the sign convention in trigonometry, an angle generated by a ray rotating in the clockwise direction is taken as negative, not positive.
Answer: False
📘 14. Very short Answer : (V.S.A.)
(C) Fill in the blanksExplanation:
By the fundamental theorem of circular measure, 1 radian is a constant angle. Therefore, any multiple of a radian, including \( \pi \) radian, represents a fixed geometric measure, making it a constant angle. Additionally, since \( \pi \text{ radian} = 180^\circ \), it forms a straight line and is also referred to as a straight angle.
Answer: constant (or straight)
Explanation:
We know that \( \pi \text{ radians} = 180^\circ \). Therefore, 1 radian = \( \left(\frac{180}{\pi}\right)^\circ \).
Using the common approximation \( \pi \approx \frac{22}{7} \):
\[\begin{array}{l} 1 \text{ radian} = \left(\frac{180 \times 7}{22}\right)^\circ = \left(\frac{630}{11}\right)^\circ \\ = 57^\circ \text{ with a remainder of } \frac{3}{11}^\circ \end{array}\]Converting the remainder to minutes (\( 1^\circ = 60' \)):
\[\begin{array}{l} \frac{3}{11} \times 60' = \frac{180'}{11} = 16' \text{ with a remainder of } \frac{4}{11}' \end{array}\]Converting the remainder to seconds (\( 1' = 60'' \)):
\[\begin{array}{l} \frac{4}{11} \times 60'' = \frac{240''}{11} \approx 22'' \end{array}\]Answer: 57° 16′ 22″
Explanation:
Two angles are supplementary if their sum is \( 180^\circ \), which is exactly \( \pi \) radians in the circular system.
Therefore, to find the supplementary angle of \( \frac{3\pi}{8} \), we subtract it from \( \pi \):
\[\begin{array}{l} \text{Supplementary angle} = \pi - \frac{3\pi}{8} \\ = \frac{8\pi}{8} - \frac{3\pi}{8} \\ = \frac{5\pi}{8} \end{array}\]Note: If the question intended "complementary" (পূরক) instead of supplementary, the sum is \( \frac{\pi}{2} \), making the answer \( \frac{\pi}{2} - \frac{3\pi}{8} = \frac{\pi}{8} \).
Answer: \( \frac{5\pi}{8} \)
📘 15. Very short Answer : (V.S.A.)
TrigonometryExplanation:
We know the standard relationship between degrees and radians is:
\[\begin{array}{l} 180^\circ = \pi \text{ radians} \end{array}\]Therefore, for any angle, the ratio of its value in degrees (D) to \( 180 \) is equal to the ratio of its value in radians (R) to \( \pi \):
\[\begin{array}{l} \frac{D}{180} = \frac{R}{\pi} \end{array}\]Rearranging the terms to find \( \frac{R}{D} \):
\[\begin{array}{l} \frac{R}{D} = \frac{\pi}{180} \end{array}\]Answer: The value of \( \frac{R}{D} \) is \( \frac{\pi}{180} \).
Explanation:
Two angles are complementary if their sum is \( 90^\circ \).
To find the complementary angle, we subtract the given angle from \( 90^\circ \).
First, let's write \( 90^\circ \) in terms of degrees, minutes, and seconds:
\[\begin{array}{l} 90^\circ = 89^\circ \, 60' \\ = 89^\circ \, 59' \, 60'' \end{array}\]Now, perform the subtraction:
\[\begin{array}{r@{\quad}l} 89^\circ \, 59' \, 60'' \\ - 63^\circ \, 35' \, 15'' \\ \hline 26^\circ \, 24' \, 45'' \end{array}\]Answer: The complementary angle is 26° 24′ 45″.
Explanation:
First, find the sum of the two given angles:
\[\begin{array}{r@{\quad}l} 65^\circ \, 56' \, 55'' \\ + 64^\circ \, 03' \, 05'' \\ \hline 129^\circ \, 59' \, 60'' \end{array}\]We know that \( 60'' = 1' \), so:
\[\begin{array}{l} 129^\circ \, 59' \, 60'' = 129^\circ \, 60' \end{array}\]And since \( 60' = 1^\circ \), we get:
\[\begin{array}{l} 129^\circ \, 60' = 130^\circ \end{array}\]The sum of the angles in a triangle is \( 180^\circ \). Let the third angle be \( x \).
\[\begin{array}{l} x = 180^\circ - 130^\circ = 50^\circ \end{array}\]Now, convert \( 50^\circ \) to circular measure (radians):
\[\begin{array}{l} \text{Circular value} = 50 \times \frac{\pi}{180} = \frac{5\pi}{18} \end{array}\]Answer: The circular value of the third angle is \( \frac{5\pi^c}{18} \).
Explanation:
We use the arc length formula: \( S = r\theta \), where \( \theta \) must be in radians.
Given:
Arc length \( (S) = 220 \) cm
Central angle in degrees \( = 63^\circ \)
Convert the angle to radians:
\[\begin{array}{l} \theta = 63 \times \frac{\pi}{180} \text{ radians} \end{array}\]Now substitute into the formula to find the radius \( r \):
\[\begin{array}{l} 220 = r \times \left(63 \times \frac{\pi}{180}\right) \\ \Rightarrow r = \frac{220 \times 180}{63 \times \pi} \end{array}\]Using \( \pi = \frac{22}{7} \):
\[\begin{array}{l} r = \frac{220 \times 180 \times 7}{63 \times 22} \\ r = \frac{10 \times 180 \times 7}{63} \quad \text{[Since } \frac{220}{22} = 10 \text{]} \\ r = \frac{10 \times 180}{9} \quad \text{[Since } \frac{7}{63} = \frac{1}{9} \text{]} \\ r = 10 \times 20 \\ r = 200 \text{ cm} \end{array}\]Answer: The radius of the circle is 200 cm.
Explanation:
The hour hand of a clock completes one full revolution (\( 360^\circ \) or \( 2\pi \) radians) in 12 hours.
Therefore, the angle formed in 1 hour is:
\[\begin{array}{l} \text{Angle in degrees} = \frac{360^\circ}{12} = 30^\circ \end{array}\]Now, convert \( 30^\circ \) to circular measure (radians):
\[\begin{array}{l} \text{Circular value} = 30 \times \frac{\pi}{180} = \frac{\pi}{6} \end{array}\]Note: Because the rotation is clockwise, the angle generated is technically negative (\( -\frac{\pi}{6} \)), but typically the magnitude is requested unless direction is explicitly asked.
Answer: The circular value of the angle is \( \frac{\pi^c}{6} \).
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